This chapter provides comprehensive ICSE Class 10 Maths Matrices Solutions from the Selina Concise Mathematics textbook. Matrices are a fundamental concept in algebra, representing data or numbers in a rectangular array of rows and columns. You will learn about the order of a matrix, different types of matrices like square, null, and identity matrices, and how to perform basic operations. We will cover the rules for adding and subtracting matrices, multiplying a matrix by a scalar, and the crucial process of multiplying two matrices together. Understanding these operations is key to solving systems of linear equations and for more advanced topics in mathematics.
If you are stuck on a specific question about matrix equality, addition, or multiplication, you’ve come to the right place. This page contains detailed, step-by-step solutions for all 81 questions found in Exercise 9(A), Exercise 9(B), Exercise 9(C), the Test Yourself section, and the Case-Study Based Question. Each solution is worked out using the exact method and format that the ICSE board expects in your exams. Here you will find clear, accurate, and reliable solutions to help you master the Matrices chapter and verify your own work.
Exercise 9(A)
Question 1(a)
If \begin{bmatrix} x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix} = \begin{bmatrix} 4 & b - 3 \\ 4 & 3 \end{bmatrix}, the value of x, y, a and b are :
- (a) x = 2, y = 1, a = 5 and b = 10
- (b) x = -2, y = 1, a = 5 and b = 10
- (c) x = 2, y = -1, a = 5 and b = 10
- (d) x = 2, y = 1, a = -5 and b = 10
We have the matrices:
\begin{bmatrix} x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix} = \begin{bmatrix} 4 & b - 3 \\ 4 & 3 \end{bmatrix}.
By comparing the corresponding elements of these matrices, we can set up the following equations:
For the first row, first column: x + 2 = 4. Solving for x, we get:
x = 4 - 2 = 2.For the second row, first column: y + 3 = 4. Solving for y, we find:
y = 4 - 3 = 1.For the first row, second column: b - 3 = 7. Solving for b, we have:
b = 7 + 3 = 10.For the second row, second column: a - 2 = 3. Solving for a, we get:
a = 3 + 2 = 5.
Notice that all the values satisfy their respective equations. Hence, option 1 is the correct option.
Question 1(b)
If A = \begin{bmatrix} 5 & -5 \\ 3 & -3 \end{bmatrix} \text{ and } B = \begin{bmatrix} -5 & 5 \\ -3 & 3 \end{bmatrix}; the value of matrix (A – B) is :
- (a) \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
- (b) \begin{bmatrix} 10 & -10 \\ 6 & -6 \end{bmatrix}
- (c) \begin{bmatrix} 10 & -10 \\ -6 & 6 \end{bmatrix}
- (d) \begin{bmatrix} -10 & 10 \\ -6 & 6 \end{bmatrix}
We are given matrices:
A = \begin{bmatrix} 5 & -5 \\ 3 & -3 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} -5 & 5 \\ -3 & 3 \end{bmatrix}To find the matrix A - B, subtract corresponding elements of matrix B from matrix A:
A - B = \begin{bmatrix} 5 & -5 \\ 3 & -3 \end{bmatrix} - \begin{bmatrix} -5 & 5 \\ -3 & 3 \end{bmatrix} = \begin{bmatrix} 5 - (-5) & -5 - 5 \\ 3 - (-3) & -3 - 3 \end{bmatrix}Calculate each element:
– First row, first column: 5 - (-5) = 5 + 5 = 10.
– First row, second column: -5 - 5 = -10.
– Second row, first column: 3 - (-3) = 3 + 3 = 6.
– Second row, second column: -3 - 3 = -6.
Thus, the resulting matrix is:
\begin{bmatrix} 10 & -10 \\ 6 & -6 \end{bmatrix}Hence, Option 2 is the correct option.
Question 1(c)
If A = \begin{bmatrix} 5 & 5 \\ 4 & 0 \end{bmatrix}, B = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \text{ and } C = \begin{bmatrix} -2 & 3 \\ 2 & 1 \end{bmatrix} then matrix (A + B – C) is :
- (a) \begin{bmatrix} 10 & 4 \\ -3 & 3 \end{bmatrix}
- (b) \begin{bmatrix} -10 & 4 \\ 3 & -3 \end{bmatrix}
- (c) \begin{bmatrix} 10 & 4 \\ 3 & 3 \end{bmatrix}
- (d) \begin{bmatrix} 10 & -4 \\ 3 & 3 \end{bmatrix}
We have matrices:
A = \begin{bmatrix} 5 & 5 \\ 4 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}, \quad C = \begin{bmatrix} -2 & 3 \\ 2 & 1 \end{bmatrix}To find A + B - C, we compute:
A + B = \begin{bmatrix} 5 & 5 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 5 + 3 & 5 + 2 \\ 4 + 1 & 0 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 5 & 4 \end{bmatrix}Now, subtract matrix C:
A + B - C = \begin{bmatrix} 8 & 7 \\ 5 & 4 \end{bmatrix} - \begin{bmatrix} -2 & 3 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 8 - (-2) & 7 - 3 \\ 5 - 2 & 4 - 1 \end{bmatrix}Simplifying each element, we get:
\begin{bmatrix} 8 + 2 & 7 - 3 \\ 5 - 2 & 4 - 1 \end{bmatrix} = \begin{bmatrix} 10 & 4 \\ 3 & 3 \end{bmatrix}Thus, the resulting matrix is \begin{bmatrix} 10 & 4 \\ 3 & 3 \end{bmatrix}.
Hence, Option 3 is the correct option.
Question 1(d)
If A = \begin{bmatrix} 7 & 5 \\ -3 & 3 \end{bmatrix} \text{ and B} = \begin{bmatrix} -2 & 5 \\ 1 & 0 \end{bmatrix}, then the matrix P (such that A + P = B) is :
- (a) \begin{bmatrix} 4 & 0 \\ 9 & -3 \end{bmatrix}
- (b) \begin{bmatrix} 9 & 0 \\ 4 & -2 \end{bmatrix}
- (c) \begin{bmatrix} -9 & 0 \\ 4 & 3 \end{bmatrix}
- (d) \begin{bmatrix} -9 & 0 \\ 4 & -3 \end{bmatrix}
We know that for matrices A and P such that A + P = B, we can find P by rearranging the equation to P = B – A.
Now, substituting the given matrices:
\Rightarrow P = \begin{bmatrix} -2 & 5 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 7 & 5 \\ -3 & 3 \end{bmatrix}Perform the subtraction element-wise:
\Rightarrow P = \begin{bmatrix} -2 - 7 & 5 - 5 \\ 1 - (-3) & 0 - 3 \end{bmatrix}Simplifying each element, we get:
\Rightarrow P = \begin{bmatrix} -9 & 0 \\ 4 & -3 \end{bmatrix}.Hence, Option 4 is the correct option.
Question 1(e)
The additive inverse of matrix A + B, where
A = \begin{bmatrix} 4 & 2 \\ 7 & -2 \end{bmatrix} \text{ and } B = \begin{bmatrix} -2 & 1 \\ 3 & -4 \end{bmatrix} is :
- (a) \begin{bmatrix} -2 & -3 \\ -10 & 6 \end{bmatrix}
- (b) \begin{bmatrix} 2 & 3 \\ -10 & -6 \end{bmatrix}
- (c) \begin{bmatrix} -2 & -3 \\ -10 & -6 \end{bmatrix}
- (d) \begin{bmatrix} -2 & 3 \\ 10 & -6 \end{bmatrix}
To find the additive inverse of a matrix, we take the negative of that matrix. Therefore, the additive inverse of the matrix (A + B) is given by -(A + B).
First, compute the sum of matrices A and B:
A = \begin{bmatrix} 4 & 2 \\ 7 & -2 \end{bmatrix} and B = \begin{bmatrix} -2 & 1 \\ 3 & -4 \end{bmatrix}.
Adding these matrices, we have:
A + B = \begin{bmatrix} 4 + (-2) & 2 + 1 \\ 7 + 3 & -2 + (-4) \end{bmatrix}.
This results in:
A + B = \begin{bmatrix} 2 & 3 \\ 10 & -6 \end{bmatrix}.
Now, to find the additive inverse, multiply each element of this matrix by -1:
-(A + B) = -\Big(\begin{bmatrix} 2 & 3 \\ 10 & -6 \end{bmatrix}\Big).
This gives us:
-(A + B) = \begin{bmatrix} -2 & -3 \\ -10 & 6 \end{bmatrix}.
Notice that this matches with option (a). Hence, option 1 is the correct option.
Question 2
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 × 2 and 2 × 3 respectively; then their sum A + B is possible.
(ii) The matrices A~2 × 3 and B~2 × 3 are conformable for subtraction.
(iii) Transpose of a 2 × 1 matrix is a 2 × 1 matrix.
(iv) Transpose of a square matrix is a square matrix.
(v) A column matrix has many columns and only one row.
(i) False
Here, matrices A and B have orders 3 × 2 and 2 × 3, which are different.
∴ The statement is false as their orders do not match.
(ii) True
Here, both matrices A and B have the order 2 × 3.
∴ The statement is true.
(iii) False
Thus, a 2 × 1 matrix becomes 1 × 2 when transposed.
∴ The statement is false.
(iv) True
∴ The statement is true.
(v) False
∴ The statement is false.
Question 3(i)
Solve for a, b and c; if :
\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}We are given the matrices \begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} and \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix} as equal. For matrices to be equal, each corresponding element must be equal.
From the first element, we have:
-4 = b + 4Solving for b, we subtract 4 from both sides:
b = -4 - 4 = -8Next, consider the second element of the first row:
a + 5 = 2Subtracting 5 from both sides gives:
a = 2 - 5 = -3Finally, for the second element of the second row:
2 = c - 1Adding 1 to both sides results in:
c = 2 + 1 = 3Hence, a = -3, b = -8 and c = 3.
Question 3(ii)
Solve for a, b and c; if :
\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}We have the matrix equation:
\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}According to the properties of matrix equality, corresponding elements must be equal. Therefore, we have:
For the element in the first row, first column:
a = 3For the element in the first row, second column:
a - b = -1
Substituting the value of a from above, we get:
3 - b = -1
Solving for b, we find:
b = 3 + 1 = 4For the element in the second row, first column:
b + c = 2
Substituting the value of b, we have:
4 + c = 2
Solving for c, we obtain:
c = -2
Hence, a = 3, b = 4, and c = -2.
Question 4
Wherever possible, write each of the following as a single matrix.
(i) \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix}
(ii) \begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix}
(iii) \begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}
(i) Consider the matrices:
\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix}To find the sum, add corresponding elements:
\Rightarrow \begin{bmatrix} 1 + (-1) & 2 + (-2) \\ 3 + 1 & 4 + (-7) \end{bmatrix}Simplifying gives:
\Rightarrow \begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.
Thus, the resulting matrix is \begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.
(ii) Consider the matrices:
\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix}Subtract corresponding elements:
\Rightarrow \begin{bmatrix} 2 - 0 & 3 - 2 & 4 - 3 \\ 5 - 6 & 6 - (-1) & 7 - 0 \end{bmatrix}Simplifying gives:
\Rightarrow \begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.
Thus, the resulting matrix is \begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.
(iii) Consider the matrices:
\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}Addition is not feasible here because the matrices differ in order.
Question 5(i)
Find, x and y from the following equations :
\begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}We start with the equation:
\Rightarrow \begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}Subtract corresponding elements in the matrices:
\Rightarrow \begin{bmatrix} 5 - 1 & 2 - (x - 1) \\ -1 - 2 & y - 1 - (-3) \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}This simplifies to:
\Rightarrow \begin{bmatrix} 4 & 3 - x \\ -3 & y + 2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}For two matrices to be equal, their corresponding elements must be identical. Thus, we have:
- For the top-right element: 3 - x = 7
Solving gives: x = 3 - 7 = -4.
For the bottom-right element: y + 2 = 2
- Solving gives: y = 0.
Thus, the values are x = -4 and y = 0.
Question 5(ii)
Find, x and y from the following equations :
\begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}We have the matrix equation:
\begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}Adding the corresponding elements from the matrices on the left-hand side, we get:
\begin{bmatrix} -8 + y & x + (-2) \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}According to the equality of matrices, the corresponding elements must be equal. Thus, we have two equations:
- -8 + y = -3
- x - 2 = 2
Let’s solve these equations one by one.
For the first equation, -8 + y = -3:
⇒ y = -3 + 8
⇒ y = 5
For the second equation, x - 2 = 2:
⇒ x = 2 + 2
⇒ x = 4
Thus, the values are x = 4 and y = 5.
Question 6
Given : M = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}, find its transpose matrix M^t. If possible, find :
(i) M + M^t
(ii) M^t – M
Consider the matrix M = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}. To find the transpose, M^t, we swap the rows and columns, resulting in M^t = \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix}.
(i) Let’s compute M + M^t:
M + M^t = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 5 + 5 & -3 + (-2) \\ -2 + (-3) & 4 + 4 \end{bmatrix} = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.Thus, M + M^t = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.
(ii) Now, calculate M^t - M:
M^t - M = \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 5 - 5 & -2 - (-3) \\ -3 - (-2) & 4 - 4 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.Therefore, M^t - M = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.
Question 7
Given A = \begin{bmatrix} 2 & -3 \end{bmatrix}, B = \begin{bmatrix} 0 & 2 \end{bmatrix} \text{ and C} = \begin{bmatrix} -1 & 4 \end{bmatrix}; find the matrix X in each of the following :
(i) X + B = C – A
(ii) A – X = B + C
(i) Starting with the equation:
∴ X + B = C – A
Rearranging gives:
⇒ X = C – A – B
Substitute the matrices A, B, and C:
\Rightarrow X = \begin{bmatrix} -1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} 0 & 2 \end{bmatrix}Perform the subtraction:
= \begin{bmatrix} -1 - 2 - 0 & 4 - (-3) - 2 \end{bmatrix}Simplify the calculations:
= \begin{bmatrix} -3 & 5 \end{bmatrix}.
Thus, the matrix X is \begin{bmatrix} -3 & 5 \end{bmatrix}.
(ii) For the second equation:
∴ A – X = B + C
Rearranging gives:
⇒ X = A – (B + C)
Substitute the matrices A, B, and C:
\Rightarrow X = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 4 \end{bmatrix}\Big)First, add matrices B and C:
= \begin{bmatrix} 0 + (-1) & 2 + 4 \end{bmatrix}This results in:
= \begin{bmatrix} -1 & 6 \end{bmatrix}Now, subtract this result from matrix A:
= \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} -1 & 6 \end{bmatrix}Perform the subtraction:
= \begin{bmatrix} 2 - (-1) & -3 - 6 \end{bmatrix}Simplify the calculations:
= \begin{bmatrix} 3 & -9 \end{bmatrix}.
Thus, the matrix X is \begin{bmatrix} 3 & -9 \end{bmatrix}.
Question 8
Given A = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}; find the matrix X in each of the following :
(i) A + X = B
(ii) A – X = B
(iii) X – B = A
(i) We start with the equation:
⇒ A + X = B
To isolate X, subtract A from both sides:
⇒ X = B – A
Now, perform the subtraction:
\Rightarrow X = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}Calculate each element:
= \begin{bmatrix} 3 - (-1) & -3 - 0 \\ -2 - 2 & 0 - (-4) \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.Thus, X is \begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.
(ii) Consider the equation:
⇒ A – X = B
To solve for X, add X to B and subtract B from A:
⇒ X = A – B
Perform the subtraction:
\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} - \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}Compute each element:
= \begin{bmatrix} -1 - 3 & 0 - (-3) \\ 2 - (-2) & -4 - 0 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.So, X is \begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.
(iii) For the equation:
⇒ X – B = A
Add B to both sides to find X:
⇒ X = A + B
Perform the addition:
\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}Calculate each element:
= \begin{bmatrix} -1 + 3 & 0 + (-3) \\ 2 + (-2) & -4 + 0 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.Hence, X is \begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.
Exercise 9(B)
Question 1(a)
If 4\begin{bmatrix} 5 & x \end{bmatrix} - 5\begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} 10 & 22 \end{bmatrix}, the values of x and y are :
- (a) x = 2 and y = 3
- (b) x = 3 and y = 2
- (c) x = -3 and y = 2
- (d) x = 3 and y = -2
We start with the equation:
4\begin{bmatrix} 5 & x \end{bmatrix} - 5\begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} 10 & 22 \end{bmatrix}This simplifies to:
\begin{bmatrix} 20 & 4x \end{bmatrix} - \begin{bmatrix} 5y & -10 \end{bmatrix} = \begin{bmatrix} 10 & 22 \end{bmatrix}Notice that we can rewrite it as:
\begin{bmatrix} 20 - 5y & 4x - (-10) \end{bmatrix} = \begin{bmatrix} 10 & 22 \end{bmatrix}Which simplifies further to:
\begin{bmatrix} 20 - 5y & 4x + 10 \end{bmatrix} = \begin{bmatrix} 10 & 22 \end{bmatrix}From here, we equate the corresponding elements:
20 - 5y = 10 \quad \text{and} \quad 4x + 10 = 22Solving these, we get:
5y = 20 - 10 \quad \text{and} \quad 4x = 22 - 10 5y = 10 \quad \text{and} \quad 4x = 12Thus, solving for y and x:
y = 2 \quad \text{and} \quad x = 3Hence, Option 2 is the correct option.
Question 1(b)
If A = \begin{bmatrix} -3 & -7 \\ 0 & -8 \end{bmatrix}\text{ and } A - B = \begin{bmatrix} 6 & 4 \\ -3 & 0 \end{bmatrix}, then matrix B is :
- (a) \begin{bmatrix} 9 & 11 \\ -3 & 18 \end{bmatrix}
- (b) \begin{bmatrix} -9 & -11 \\ 3 & 8 \end{bmatrix}
- (c) \begin{bmatrix} 9 & -11 \\ -3 & 8 \end{bmatrix}
- (d) \begin{bmatrix} -9 & -11 \\ -3 & -8 \end{bmatrix}
We have the equation:
A – B = \begin{bmatrix} 6 & 4 \\ -3 & 0 \end{bmatrix}
By substituting the given matrix A into this equation, we get:
\Rightarrow \begin{bmatrix} -3 & -7 \\ 0 & -8 \end{bmatrix} - B = \begin{bmatrix} 6 & 4 \\ -3 & 0 \end{bmatrix}To find matrix B, rearrange the equation:
\Rightarrow B = \begin{bmatrix} -3 & -7 \\ 0 & -8 \end{bmatrix} - \begin{bmatrix} 6 & 4 \\ -3 & 0 \end{bmatrix}Now, perform the subtraction element-wise:
\Rightarrow B = \begin{bmatrix} -3 - 6 & -7 - 4 \\ 0 - (-3) & -8 - 0 \end{bmatrix}Simplifying each element, we find:
\Rightarrow B = \begin{bmatrix} -9 & -11 \\ 3 & -8 \end{bmatrix}Hence, Option 4 is the correct option.
Question 1(c)
If I is a unit matrix of order 2 and M + 4I = \begin{bmatrix} 8 & -3 \\ 4 & 2 \end{bmatrix}, then matrix M is :
- (a) \begin{bmatrix} 4 & 3 \\ 4 & -2 \end{bmatrix}
- (b) \begin{bmatrix} 4 & 3 \\ 4 & 2 \end{bmatrix}
- (c) \begin{bmatrix} 4 & -3 \\ -4 & 2 \end{bmatrix}
- (d) \begin{bmatrix} 4 & -3 \\ 4 & -2 \end{bmatrix}
The unit matrix I of order 2 is given by:
I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}We know:
M + 4I = \begin{bmatrix} 8 & -3 \\ 4 & 2 \end{bmatrix}Substituting the unit matrix I, we have:
M + 4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -3 \\ 4 & 2 \end{bmatrix}This simplifies to:
M + \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 8 & -3 \\ 4 & 2 \end{bmatrix}To find M, subtract \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} from \begin{bmatrix} 8 & -3 \\ 4 & 2 \end{bmatrix}:
M = \begin{bmatrix} 8 & -3 \\ 4 & 2 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}Calculate the result:
M = \begin{bmatrix} 8 - 4 & -3 - 0 \\ 4 - 0 & 2 - 4 \end{bmatrix}Thus, we get:
M = \begin{bmatrix} 4 & -3 \\ 4 & -2 \end{bmatrix}Hence, Option 4 is the correct option.
Question 1(d)
If 2\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}, the values of x, y and z are :
- (a) x = 8, y = -5 and z = 9
- (b) x = -8, y = 5 and z = 9
- (c) x = -8, y = -5 and z = -9
- (d) x = -8, y = 5 and z = -9
We’re given the equation:
2\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}First, multiply each matrix by its scalar:
\begin{bmatrix} 6 & 2x \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ 3y & 6 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}Now, add the corresponding elements of the matrices:
\begin{bmatrix} 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}This simplifies to:
\begin{bmatrix} 9 & 2x + 9 \\ 3y & 8 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}From this, we equate the corresponding elements:
- For the first element, z = 9
- For the second element, 2x + 9 = -7
- For the third element, 3y = 15
Solving these, we find:
- z = 9
- y = \frac{15}{3} = 5
- 2x = -7 - 9 = -16
Thus, solving for x:
- x = \frac{-16}{2} = -8
∴ We find that z = 9, y = 5, and x = -8. Hence, Option 2 is the correct option.
Question 1(e)
Given A = \begin{bmatrix} 4 & 7 \\ 3 & -2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}, then A – 2B is :
- (a) \begin{bmatrix} -2 & 3 \\ 5 & -10 \end{bmatrix}
- (b) \begin{bmatrix} -2 & -3 \\ -5 & 10 \end{bmatrix}
- (c) \begin{bmatrix} 2 & 3 \\ 5 & -10 \end{bmatrix}
- (d) \begin{bmatrix} 2 & 3 \\ 5 & 10 \end{bmatrix}
To find A - 2B, we start by calculating 2B. Multiply each element of matrix B = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} by 2:
2B = 2 \times \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times (-1) & 2 \times 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ -2 & 8 \end{bmatrix}Next, subtract 2B from A = \begin{bmatrix} 4 & 7 \\ 3 & -2 \end{bmatrix}:
A - 2B = \begin{bmatrix} 4 & 7 \\ 3 & -2 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ -2 & 8 \end{bmatrix} = \begin{bmatrix} 4 - 2 & 7 - 4 \\ 3 - (-2) & -2 - 8 \end{bmatrix}Perform the subtraction for each corresponding element:
= \begin{bmatrix} 2 & 3 \\ 3 + 2 & -10 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 5 & -10 \end{bmatrix}Thus, the resulting matrix is \begin{bmatrix} 2 & 3 \\ 5 & -10 \end{bmatrix}. Option 3 is the correct option.
Question 2
Find x and y if :
(i) 3\begin{bmatrix} 4 & x \end{bmatrix} + 2\begin{bmatrix} y & -3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix}
(ii) x\begin{bmatrix} -1 \\ 2 \end{bmatrix} - 4\begin{bmatrix} -2 \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}
(i) Consider the equation:
3\begin{bmatrix} 4 & x \end{bmatrix} + 2\begin{bmatrix} y & -3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix}This simplifies to:
\begin{bmatrix} 12 & 3x \end{bmatrix} + \begin{bmatrix} 2y & -6 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix}Combining the matrices gives:
\begin{bmatrix} 12 + 2y & 3x - 6 \end{bmatrix} = \begin{bmatrix} 10 & 0 \end{bmatrix}By equating corresponding elements, we have:
\begin{aligned}12 + 2y = 10 \\3x - 6 = 0\end{aligned}Solving the first equation:
\begin{aligned}2y = 10 - 12 \\2y = -2 \\y = -1\end{aligned}And for the second equation:
\begin{aligned}3x = 6 \\x = 2\end{aligned}Thus, x = 2 and y = -1.
(ii) Now, consider:
x\begin{bmatrix} -1 \\ 2 \end{bmatrix} - 4\begin{bmatrix} -2 \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}This becomes:
\begin{bmatrix} -x \\ 2x \end{bmatrix} - \begin{bmatrix} -8 \\ 4y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}Simplifying, we get:
\begin{bmatrix} -x + 8 \\ 2x - 4y \end{bmatrix} = \begin{bmatrix} 7 \\ -8 \end{bmatrix}Equating the elements, we derive:
\begin{aligned}-x + 8 = 7 \\2x - 4y = -8\end{aligned}From the first equation:
\begin{aligned}x = 8 - 7 \\x = 1\end{aligned}Substituting x = 1 into the second equation:
\begin{aligned}2(1) - 4y = -8 \\2 - 4y = -8 \\4y = 2 + 8 \\4y = 10 \\y = \dfrac{10}{4} = \dfrac{5}{2} = 2.5\end{aligned}Thus, x = 1 and y = 2.5.
Question 3
Given A = \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}, B = \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} \text{ and } C = \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}; find :
(i) 2A – 3B + C
(ii) A + 2C – B
(i) We need to calculate 2A – 3B + C. Begin by substituting the given matrices A, B, and C into the expression:
2A - 3B + C = 2\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} - 3\begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} + \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}Calculate each term separately:
2 \times \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix} 3 \times \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 15 & 6 \end{bmatrix}Now, substitute back and simplify:
\begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix} - \begin{bmatrix} 3 & 3 \\ 15 & 6 \end{bmatrix} + \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 4 - 3 + (-3) & 2 - 3 + (-1) \\ 6 - 15 + 0 & 0 - 6 + 0 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -9 & -6 \end{bmatrix}∴ 2A – 3B + C = \begin{bmatrix}[r] -2 & -2 \ -9 & -6 \end{bmatrix}.
(ii) Now, let’s find A + 2C – B. Substitute the matrices into the expression:
A + 2C - B = \begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} + 2\begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}Calculate the multiplication:
2 \times \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} -6 & -2 \\ 0 & 0 \end{bmatrix}Substitute back and simplify:
\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix} + \begin{bmatrix} -6 & -2 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 2 + (-6) - 1 & 1 + (-2) - 1 \\ 3 + 0 - 5 & 0 + 0 - 2 \end{bmatrix} = \begin{bmatrix} -5 & -2 \\ -2 & -2 \end{bmatrix}∴ A + 2C – B = \begin{bmatrix}[r] -5 & -2 \ -2 & -2 \end{bmatrix}.
Question 4
If \begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix} + 3A = \begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix}; find A.
We start with the equation:
\begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix} + 3A = \begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix}To isolate 3A, subtract \begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix} from both sides:
3A = \begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix}Perform the subtraction element-wise:
3A = \begin{bmatrix} -2 - 4 & -2 - (-2) \\ 1 - 4 & -3 - 0 \end{bmatrix}This simplifies to:
3A = \begin{bmatrix} -6 & 0 \\ -3 & -3 \end{bmatrix}To find A, divide each element of the matrix by 3:
A = \dfrac{1}{3}\begin{bmatrix} -6 & 0 \\ -3 & -3 \end{bmatrix}Thus, we obtain:
A = \begin{bmatrix} -2 & 0 \\ -1 & -1 \end{bmatrix}Therefore, the matrix A is \begin{bmatrix} -2 & 0 \\ -1 & -1 \end{bmatrix}.
Question 5
Given A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} \text{ and } B = \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}
(i) find the matrix 2A + B
(ii) find a matrix C such that :
C + B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
(i)
To find 2A + B, we start by calculating 2A. Multiply each element of matrix A by 2:
2 \times \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ 4 & 6 \end{bmatrix}.
Now, add matrix B to 2A:
\begin{bmatrix} 2 & 8 \\ 4 & 6 \end{bmatrix} + \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix} = \begin{bmatrix} 2 + (-4) & 8 + (-1) \\ 4 + (-3) & 6 + (-2) \end{bmatrix}.
This results in:
\begin{bmatrix} -2 & 7 \\ 1 & 4 \end{bmatrix}.
Thus, 2A + B = \begin{bmatrix} -2 & 7 \\ 1 & 4 \end{bmatrix}.
(ii)
To find matrix C, we know:
C + B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.
Rearrange to solve for C:
C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - B.
Substitute matrix B:
C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}.
Calculate each element:
C = \begin{bmatrix} 0 - (-4) & 0 - (-1) \\ 0 - (-3) & 0 - (-2) \end{bmatrix}.
This simplifies to:
\begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}.
Thus, C = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}.
Question 6
If 2\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}; find the values of x, y and z.
We have the matrix equation:
2\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} + 3\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}Start by multiplying each matrix by its scalar:
2\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 2x \\ 0 & 2 \end{bmatrix}and
3\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 3y & 6 \end{bmatrix}Now, add the resulting matrices:
\begin{bmatrix} 6 & 2x \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ 3y & 6 \end{bmatrix} = \begin{bmatrix} 6+3 & 2x+9 \\ 0+3y & 2+6 \end{bmatrix} = \begin{bmatrix} 9 & 2x+9 \\ 3y & 8 \end{bmatrix}This matrix is equal to \begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}. According to the equality of matrices, each corresponding element must be equal:
- For the first row, first column: z = 9
- For the first row, second column: 2x + 9 = -7
- Solving for x:
2x = -7 - 9
2x = -16
x = -8 - For the second row, first column: 3y = 15
- Solving for y:
y = 5
Thus, the values are x = -8, y = 5, and z = 9.
Question 7
Given A = \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} and A^t is its transpose matrix. Find :
(i) 2A + 3A^t
(ii) 2A^t – 3A
(iii) \dfrac{1}{2}A - \dfrac{1}{3}A^t
(iv) A^t - \dfrac{1}{3}A
A = \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} and A^t = \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix}.
(i) To compute 2A + 3A^t, substitute the matrices A and A^t:
\Rightarrow 2A + 3A^t = 2\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} + 3\begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} \\= \begin{bmatrix} -6 & 12 \\ 0 & -18 \end{bmatrix} + \begin{bmatrix} -9 & 0 \\ 18 & -27 \end{bmatrix} \\= \begin{bmatrix} -6 + (-9) & 12 + 0 \\ 0 + 18 & -18 + (-27) \end{bmatrix} \\= \begin{bmatrix} -15 & 12 \\ 18 & -45 \end{bmatrix}.Thus, 2A + 3A^t = \begin{bmatrix} -15 & 12 \\ 18 & -45 \end{bmatrix}.
(ii) For 2A^t - 3A, replace A and A^t in the expression:
\Rightarrow 2A^t - 3A = 2\begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} - 3\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} \\= \begin{bmatrix} -6 & 0 \\ 12 & -18 \end{bmatrix} - \begin{bmatrix} -9 & 18 \\ 0 & -27 \end{bmatrix} \\= \begin{bmatrix} -6 - (-9) & 0 - 18 \\ 12 - 0 & -18 - (-27) \end{bmatrix} \\= \begin{bmatrix} 3 & -18 \\ 12 & 9 \end{bmatrix}.Hence, 2A^t - 3A = \begin{bmatrix} 3 & -18 \\ 12 & 9 \end{bmatrix}.
(iii) To find \dfrac{1}{2}A - \dfrac{1}{3}A^t, substitute A and A^t:
\Rightarrow \dfrac{1}{2}A - \dfrac{1}{3}A^t = \dfrac{1}{2}\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} - \dfrac{1}{3}\begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} \\= \begin{bmatrix} -\dfrac{3}{2} & 3 \\ 0 & -\dfrac{9}{2} \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 2 & -3 \end{bmatrix} \\= \begin{bmatrix} -\dfrac{3}{2} - (-1) & 3 - 0 \\ 0 - 2 & -\dfrac{9}{2} - (-3) \end{bmatrix} \\= \begin{bmatrix} -\dfrac{1}{2} & 3 \\ -2 & -\dfrac{3}{2} \end{bmatrix}Therefore, \dfrac{1}{2}A - \dfrac{1}{3}A^t = \begin{bmatrix} -\dfrac{1}{2} & 3 \\ -2 & -\dfrac{3}{2} \end{bmatrix}.
(iv) For A^t - \dfrac{1}{3}A, substitute A and A^t:
\Rightarrow A^t - \dfrac{1}{3}A = \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} - \dfrac{1}{3}\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix} \\= \begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix} - \begin{bmatrix} -1 & 2 \\ 0 & -3 \end{bmatrix} \\= \begin{bmatrix} -3 - (-1) & 0 - 2 \\ 6 - 0 & -9 - (-3) \end{bmatrix} \\= \begin{bmatrix} -2 & -2 \\ 6 & -6 \end{bmatrix}.Thus, A^t - \dfrac{1}{3}A = \begin{bmatrix} -2 & -2 \\ 6 & -6 \end{bmatrix}.
Question 8
Given A = \begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}.
Solve for matrix X :
(i) X + 2A = B
(ii) 3x + B + 2A = 0
(iii) 3A – 2X = X – 2B.
(i) We start with the equation:
X + 2A = BTo isolate X, subtract 2A from both sides:
X = B - 2ASubstitute the given matrices A and B:
X = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} - 2\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix}Calculate 2A:
2A = \begin{bmatrix} 2 & 2 \\ -4 & 0 \end{bmatrix}Now perform the subtraction:
X = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 2 \\ -4 & 0 \end{bmatrix}Calculate each element:
X = \begin{bmatrix} 2 - 2 & -1 - 2 \\ 1 - (-4) & 1 - 0 \end{bmatrix}This results in:
X = \begin{bmatrix} 0 & -3 \\ 5 & 1 \end{bmatrix}Thus, X = \begin{bmatrix} 0 & -3 \\ 5 & 1 \end{bmatrix}.
(ii) For the equation:
3X + B + 2A = 0Rearrange to find 3X:
3X = -(B + 2A)Substitute the matrices B and 2A:
3X = -\Big(\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 2 \\ -4 & 0 \end{bmatrix}\Big)Add the matrices:
3X = -\Big(\begin{bmatrix} 4 & 1 \\ -3 & 1 \end{bmatrix}\Big)Now, calculate X by dividing by 3:
X = -\dfrac{1}{3}\Big(\begin{bmatrix} 4 & 1 \\ -3 & 1 \end{bmatrix}\Big)This gives:
X = \begin{bmatrix} -\dfrac{4}{3} & -\dfrac{1}{3} \\ 1 & -\dfrac{1}{3} \end{bmatrix}Thus, X = \begin{bmatrix} -\dfrac{4}{3} & -\dfrac{1}{3} \\ 1 & -\dfrac{1}{3} \end{bmatrix}.
(iii) Consider the equation:
3A - 2X = X - 2BRearrange terms to get:
X + 2X = 3A + 2BSimplify to:
3X = 3A + 2BSubstitute the matrices A and B:
3X = 3\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} + 2\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}Calculate 3A and 2B:
3A = \begin{bmatrix} 3 & 3 \\ -6 & 0 \end{bmatrix} 2B = \begin{bmatrix} 4 & -2 \\ 2 & 2 \end{bmatrix}Add these results:
3X = \begin{bmatrix} 7 & 1 \\ -4 & 2 \end{bmatrix}Divide by 3 to find X:
X = \dfrac{1}{3}\begin{bmatrix} 7 & 1 \\ -4 & 2 \end{bmatrix}This results in:
X = \begin{bmatrix} \dfrac{7}{3} & \dfrac{1}{3} \\ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix}Thus, X = \begin{bmatrix} \dfrac{7}{3} & \dfrac{1}{3} \\ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix}.
Question 9
If I is the unit matrix of order 2 × 2; find the matrix M such that :
5M + 3I = 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix}
The unit matrix I of order 2 \times 2 is given by:
I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}We are provided with the equation:
5M + 3I = 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix}Substituting the unit matrix I, we have:
5M + 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -20 \\ 0 & -12 \end{bmatrix}This simplifies to:
5M + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & -20 \\ 0 & -12 \end{bmatrix}To isolate 5M, subtract \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} from both sides:
5M = \begin{bmatrix} 8 & -20 \\ 0 & -12 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}This results in:
5M = \begin{bmatrix} 8 - 3 & -20 - 0 \\ 0 - 0 & -12 - 3 \end{bmatrix}Simplifying further, we get:
5M = \begin{bmatrix} 5 & -20 \\ 0 & -15 \end{bmatrix}To find M, divide the entire matrix by 5:
M = \dfrac{1}{5}\begin{bmatrix} 5 & -20 \\ 0 & -15 \end{bmatrix}Thus, M becomes:
M = \begin{bmatrix} 1 & -4 \\ 0 & -3 \end{bmatrix}.
Therefore, the matrix M is \begin{bmatrix} 1 & -4 \\ 0 & -3 \end{bmatrix}.
Exercise 9(C)
Question 1(a)
If A is a matrix of order m × 3, B is a matrix of order 3 × 2 and R is a matrix of order 5 × n such that AB = R, the values of m and n are :
- (a) m = -5 and n = -2
- (b) m = 5 and n = 2
- (c) m = 5 and n = -2
- (d) m = 2 and n = 5
We have the equation AB = R, where matrix A has dimensions m × 3, matrix B has dimensions 3 × 2, and matrix R has dimensions 5 × n.
For matrix multiplication to be defined, the number of columns in matrix A must match the number of rows in matrix B. This condition is satisfied here as both are 3.
The resulting matrix from the multiplication AB will have dimensions determined by the number of rows in A and the number of columns in B. Therefore, the order of the resulting matrix R is m × 2.
Since R is given to be of order 5 × n, it follows that m = 5 and n = 2.
Hence, Option 2 is the correct option.
Question 1(b)
If A = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} and A = B^2, the value of x is :
- (a) 38
- (b) -6
- (c) -36
- (d) 36
Given matrices A and B, we know that A = B^2. Let’s calculate B^2 to find the value of x.
Substituting the matrices, we have:
\Rightarrow \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix}Now, perform the matrix multiplication on the right side:
\Rightarrow \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix}Simplifying the multiplication gives:
\Rightarrow \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix}Which simplifies to:
\Rightarrow \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}Notice that the elements in the first row and second column of both matrices must be equal, so x = 36.
Hence, Option 4 is the correct option.
Question 1(c)
A, B and C are three square matrices each of order 3; the order of matrix CA + B^2 is :
- (a) 3 × 1
- (b) 3 × 3
- (c) 1 × 3
- (d) 2 × 3
Consider the matrix multiplication for CA. Here, C is a 3 × 3 matrix and A is also a 3 × 3 matrix. The product CA will be a matrix R of order m × n.
For matrix multiplication to be valid, the number of columns in C must match the number of rows in A, which they do. Thus, the order of the resulting matrix R will be determined by the number of rows in C and the number of columns in A. Therefore, m = 3 and n = 3, making the order of matrix R 3 × 3.
Next, let’s look at the matrix B^2. Since B is a 3 × 3 matrix, multiplying B by itself will give us a matrix S of order g × h. The order of matrix S is determined by the number of rows in B and the number of columns in B. Hence, g = 3 and h = 3, so the order of matrix S is 3 × 3.
Now, for the expression CA + B^2, we have R + S. The addition of two matrices is possible only if they have the same order, which they do in this case. Therefore, the order of the resultant matrix from this addition is also 3 × 3.
Hence, Option 2 is the correct option.
Question 1(d)
If A = \begin{bmatrix} 5 & -2 \\ 7 & 0 \end{bmatrix} \text{ and B} = \begin{bmatrix} 8 \\ 3 \end{bmatrix}, then which of the following is not possible ?
- (a) A^2
- (b) AB
- (c) BA
- (d) 15A
To determine the feasibility of multiplying matrices, recall that the number of columns in the first matrix must match the number of rows in the second matrix. Here, matrix B is a 2 \times 1 matrix, meaning it has 1 column. Matrix A is a 2 \times 2 matrix, meaning it has 2 rows. Since the number of columns in B (1) does not equal the number of rows in A (2), the multiplication BA is not possible.
Hence, Option 3 is the correct option.
Question 1(e)
If A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\text{ and C} = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, the matrix A^2 + 2B – 3C is :
- (a) \begin{bmatrix} -2 & -1 \\ 4 & 1 \end{bmatrix}
- (b) \begin{bmatrix} 2 & -1 \\ 4 & 1 \end{bmatrix}
- (c) \begin{bmatrix} 2 & 1 \\ 4 & 1 \end{bmatrix}
- (d) \begin{bmatrix} 2 & 1 \\ -4 & -1 \end{bmatrix}
To determine the expression A^2 + 2B - 3C, we first need to calculate each component. Let’s start by finding A^2. Multiply matrix A by itself:
A^2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} (1 \times 1 + 0 \times 1) & (1 \times 0 + 0 \times 1) \\ (1 \times 1 + 1 \times 1) & (1 \times 0 + 1 \times 1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}.Next, calculate 2B by multiplying each element of B by 2:
2B = 2 \times \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}.Now, find 3C by multiplying each element of C by 3:
3C = 3 \times \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 0 & 0 \end{bmatrix}.Now substitute these results into the expression A^2 + 2B - 3C:
A^2 + 2B - 3C = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix} - \begin{bmatrix} 3 & 3 \\ 0 & 0 \end{bmatrix}.Add the matrices A^2 and 2B:
\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix}.Now subtract 3C from the result:
\begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 - 3 & 2 - 3 \\ 4 - 0 & 1 - 0 \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ 4 & 1 \end{bmatrix}.∴ The resulting matrix is \begin{bmatrix} -2 & -1 \\ 4 & 1 \end{bmatrix}. Hence, Option 1 is the correct option.
Question 2(i)
Evaluate : if possible :
\begin{bmatrix} 3 & 2 \\ \end{bmatrix}\begin{bmatrix} 2 \\ 0 \end{bmatrix}Consider the matrix multiplication: \begin{bmatrix} 3 & 2 \\ \end{bmatrix}\begin{bmatrix} 2 \\ 0 \end{bmatrix}.
For the multiplication of two matrices to be valid, the number of columns in the first matrix must match the number of rows in the second matrix. Here, both conditions are satisfied, so the multiplication is feasible.
Perform the multiplication as follows:
\Rightarrow \begin{bmatrix} 3 & 2 \\ \end{bmatrix}\begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \times 2 + 2 \times 0 \\ \end{bmatrix} \\ = \begin{bmatrix} 6 \\ \end{bmatrix}.Thus, the result of the matrix multiplication is \begin{bmatrix} 6 \end{bmatrix}.
Question 2(ii)
Evaluate : if possible :
\begin{bmatrix} 1 & -2 \\ \end{bmatrix}\begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix}To determine if the multiplication of matrices is possible, ensure that the number of columns in the first matrix matches the number of rows in the second matrix.
Here, we have:
\begin{bmatrix} 1 & -2 \\ \end{bmatrix}\begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix}The first matrix has 2 columns, and the second matrix has 2 rows, so multiplication is possible.
Perform the multiplication as follows:
\Rightarrow \begin{bmatrix} 1 & -2 \\ \end{bmatrix}\begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 \times (-2) + (-2) \times (-1) & 1 \times 3 + (-2) \times 4 \end{bmatrix}Calculate each element:
\begin{bmatrix} -2 + 2 & 3 + (-8) \\ \end{bmatrix}Simplifying gives:
\begin{bmatrix} 0 & -5 \\ \end{bmatrix}Thus, the resultant matrix is \begin{bmatrix} 0 & -5 \\ \end{bmatrix}.”}
Question 2(iii)
Evaluate : if possible :
\begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 3 \end{bmatrix}To multiply these matrices, ensure the number of columns in the first matrix matches the number of rows in the second matrix. Here, the first matrix is 2 \times 2 and the second matrix is 2 \times 1, so multiplication is possible.
Perform the multiplication as follows:
\Rightarrow \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 6 \times (-1) + 4 \times 3 \\ 3 \times (-1) + (-1) \times 3 \end{bmatrix}Calculate each element:
- For the first element: 6 \times (-1) + 4 \times 3 = -6 + 12 = 6.
- For the second element: 3 \times (-1) + (-1) \times 3 = -3 + (-3) = -6.
Thus, the resulting matrix is:
\begin{bmatrix} 6 \\ -6 \end{bmatrix}.
Therefore, \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 6 \\ -6 \end{bmatrix}.
Question 2(iv)
Evaluate : if possible :
\begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} -1 & 3 \\ \end{bmatrix}To multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix. Here, we have the matrix \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix} with 2 columns, and the matrix \begin{bmatrix} -1 & 3 \\ \end{bmatrix} with only 1 row. Since these numbers do not match, this multiplication cannot be performed. Thus, the matrix multiplication is not possible.
Question 3
If A = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix}, B = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} and I is a unit matrix of order 2 × 2, find :
(i) AB
(ii) BA
(iii) AI
(i) To determine the product AB, we multiply the matrices:
\Rightarrow AB = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}Let’s calculate each element of the resulting matrix:
- First row, first column: 0 \times 1 + 2 \times 3 = 0 + 6 = 6
- First row, second column: 0 \times (-1) + 2 \times 2 = 0 + 4 = 4
- Second row, first column: 5 \times 1 + (-2) \times 3 = 5 - 6 = -1
- Second row, second column: 5 \times (-1) + (-2) \times 2 = -5 - 4 = -9
Thus, AB = \begin{bmatrix} 6 & 4 \\ -1 & -9 \end{bmatrix}.
(ii) Now, let’s find BA by multiplying B and A:
\Rightarrow BA = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix}Calculate each element:
- First row, first column: 1 \times 0 + (-1) \times 5 = 0 - 5 = -5
- First row, second column: 1 \times 2 + (-1) \times (-2) = 2 + 2 = 4
- Second row, first column: 3 \times 0 + 2 \times 5 = 0 + 10 = 10
- Second row, second column: 3 \times 2 + 2 \times (-2) = 6 - 4 = 2
So, BA = \begin{bmatrix} -5 & 4 \\ 10 & 2 \end{bmatrix}.
(iii) Finally, compute AI where I is the identity matrix:
\Rightarrow AI = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Calculate the product:
- First row, first column: 0 \times 1 + 2 \times 0 = 0
- First row, second column: 0 \times 0 + 2 \times 1 = 2
- Second row, first column: 5 \times 1 + (-2) \times 0 = 5
- Second row, second column: 5 \times 0 + (-2) \times 1 = -2
Thus, AI = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} = A.
Hence, AI = matrix A.
Question 4
If A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \text{ and B }= \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}, find x and y when A^2 = B.
We start with the equation A^2 = B.
This implies:
\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}When we multiply the matrices on the left, we calculate each element as follows:
- The element in the first row, first column is 3 \times 3 + x \times 0 = 9 + 0.
- The element in the first row, second column is 3 \times x + x \times 1 = 3x + x.
- The element in the second row, first column is 0 \times 3 + 1 \times 0 = 0 + 0.
- The element in the second row, second column is 0 \times x + 1 \times 1 = 0 + 1.
This results in:
\begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}Now, comparing corresponding elements from both matrices:
- For the first row, second column: 4x = 16 \Rightarrow x = 4.
- For the second row, second column: 1 = -y \Rightarrow y = -1.
Hence, x = 4 and y = -1.
Question 5
Find x and y, if :
\begin{bmatrix} x & 0 \\ -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}We start with the given matrix equation:
\begin{bmatrix} x & 0 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}To find the product of these matrices, calculate:
\begin{bmatrix} x \times 1 + 0 \times 0 & x \times 1 + 0 \times y \\ -3 \times 1 + 1 \times 0 & -3 \times 1 + 1 \times y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}Simplifying further, we have:
\begin{bmatrix} x + 0 & x + 0 \\ -3 + 0 & -3 + y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}This reduces to:
\begin{bmatrix} x & x \\ -3 & -3 + y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}By comparing the corresponding elements of the matrices, we find:
- For the first row:
The element in the first column gives us: x = 2
For the second row:
- The element in the second column gives us: -3 + y = -2
- Solving this, we find: y = -2 + 3 = 1
Hence, x = 2 and y = 1.
Question 6
If A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, B = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}, find :
(i) (AB)C
(ii) A(BC)
Is A(BC) = (AB)C ?
(i) To compute (AB)C, substitute the matrices A, B, and C:
\Rightarrow \Big(\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\Big)\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}Calculate the product of matrices A and B:
= \Big(\begin{bmatrix} 1 \times 1 + 3 \times 4 & 1 \times 2 + 3 \times 3 \\ 2 \times 1 + 4 \times 4 & 2 \times 2 + 4 \times 3 \end{bmatrix}\Big)\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} = \Big(\begin{bmatrix} 1 + 12 & 2 + 9 \\ 2 + 16 & 4 + 12 \end{bmatrix}\Big)\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 13 & 11 \\ 18 & 16 \end{bmatrix}\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}Now, multiply the resulting matrix by C:
= \begin{bmatrix} 13 \times 4 + 11 \times 1 & 13 \times 3 + 11 \times 2 \\ 18 \times 4 + 16 \times 1 & 18 \times 3 + 16 \times 2 \end{bmatrix} = \begin{bmatrix} 52 + 11 & 39 + 22 \\ 72 + 16 & 54 + 32 \end{bmatrix}= \begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}.
∴ (AB)C = \begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}.
(ii) For A(BC), substitute the matrices A, B, and C:
\Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\Big(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\Big)Calculate the product of matrices B and C:
= \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\Big(\begin{bmatrix} 1 \times 4 + 2 \times 1 & 1 \times 3 + 2 \times 2 \\ 4 \times 4 + 3 \times 1 & 4 \times 3 + 3 \times 2 \end{bmatrix}\Big) = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} 4 + 2 & 3 + 4 \\ 16 + 3 & 12 + 6 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} 6 & 7 \\ 19 & 18 \end{bmatrix}Now, multiply matrix A by the resulting matrix:
= \begin{bmatrix} 1 \times 6 + 3 \times 19 & 1 \times 7 + 3 \times 18 \\ 2 \times 6 + 4 \times 19 & 2 \times 7 + 4 \times 18 \end{bmatrix} = \begin{bmatrix} 6 + 57 & 7 + 54 \\ 12 + 76 & 14 + 72 \end{bmatrix} = \begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}∴ A(BC) = \begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}.
Therefore, A(BC) = (AB)C.
Question 7
Let A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}, B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \text{ and C }= \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}. Find A^2 + AC – 5B.
Let’s evaluate the expression A^2 + AC - 5B using the given matrices.
First, calculate A^2 by multiplying matrix A with itself:
\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\0 \times 2 + (-2) \times 0 & 0 \times 1 + (-2) \times (-2) \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}Next, compute AC by multiplying matrices A and C:
\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \times \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 2 \times (-3) + 1 \times (-1) & 2 \times 2 + 1 \times 4 \\0 \times (-3) + (-2) \times (-1) & 0 \times 2 + (-2) \times 4 \end{bmatrix} = \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}For the term 5B, multiply matrix B by 5:
5 \times \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} = \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}Now, substitute these results into the original expression:
\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}Perform the matrix addition and subtraction:
\begin{bmatrix} 4 + (-7) - 20 & 0 + 8 - 5 \\0 + 2 - (-15) & 4 + (-8) - (-10) \end{bmatrix} = \begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix}Consequently, A^2 + AC - 5B = \begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix}.
Question 8
If M = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} and I is a unit matrix of the same order as that of M; show that :
M^2 = 2M + 3I
The identity matrix I for our case is given by \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.\
\
Given matrix M is \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}.\
\
Let’s compute the left-hand side (LHS), M^2: \
\
M^2 = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \
\
Performing matrix multiplication, we have: \
\
= \begin{bmatrix} 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \end{bmatrix} \
\
This simplifies to: \
\
= \begin{bmatrix} 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{bmatrix} \
\
= \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}.\
\
Now, consider the right-hand side (RHS), 2M + 3I: \
\
2M = 2 \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix} \
\
3I = 3 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \
\
Adding these results: \
\
2M + 3I = \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \
\
= \begin{bmatrix} 2 + 3 & 4 + 0 \\ 4 + 0 & 2 + 3 \end{bmatrix} \
\
= \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}.\
\
Notice that LHS = RHS.\
\
Hence, proved that M^2 = 2M + 3I.
Question 9
If A = \begin{bmatrix} a & 0 \\ 0 & 2 \end{bmatrix}, B = \begin{bmatrix} 0 & -b \\ 1 & 0 \end{bmatrix}, M = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} and BA = M^2, find the values of a and b.
We start by using the given condition BA = M^2. This translates to:
\begin{bmatrix} 0 & -b \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}Now, let’s compute each side. For the left-hand side, perform matrix multiplication:
\begin{bmatrix} 0 \times a + (-b) \times 0 & 0 \times 0 + (-b) \times 2 \\1 \times a + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & -2b \\ a & 0 \end{bmatrix}For the right-hand side, compute M^2:
\begin{bmatrix} 1 \times 1 + (-1) \times 1 & 1 \times (-1) + (-1) \times 1 \\1 \times 1 + 1 \times 1 & 1 \times (-1) + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}Equating the matrices from both sides, we apply the property of equality of matrices:
From the first row, second column: -2b = -2
⇒ b = 1.From the second row, first column: a = 2.
Thus, we find that a = 2 and b = 1.**
Question 10
Find the matrix A, if B = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} \text{ and } B^2 = B + \dfrac{1}{2}A.
We have the equation:
B^2 = B + \dfrac{1}{2}ASubstitute the matrix B = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} into this equation:
\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} + \dfrac{1}{2}ACalculate B^2 by performing matrix multiplication:
\begin{bmatrix} 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times 1 \\0 \times 2 + 1 \times 0 & 0 \times 1 + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} + \dfrac{1}{2}AThis simplifies to:
\begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} + \dfrac{1}{2}ASubtract B from both sides to isolate \dfrac{1}{2}A:
\dfrac{1}{2}A = \begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}Perform the subtraction:
\dfrac{1}{2}A = \begin{bmatrix} 4 - 2 & 3 - 1 \\ 0 - 0 & 1 - 1 \end{bmatrix}This results in:
\dfrac{1}{2}A = \begin{bmatrix} 2 & 2 \\ 0 & 0 \end{bmatrix}Multiply through by 2 to solve for A:
A = 2 \begin{bmatrix} 2 & 2 \\ 0 & 0 \end{bmatrix}Therefore, we find:
A = \begin{bmatrix} 4 & 4 \\ 0 & 0 \end{bmatrix}Thus, the matrix A is \begin{bmatrix} 4 & 4 \\ 0 & 0 \end{bmatrix}.
Question 11
If A = \begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix} and A^2 = I, find a and b.
We know that A^2 = I where I is the identity matrix. So we have:
\begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix}\begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Carrying out the matrix multiplication on the left-hand side, we get:
\begin{bmatrix} (-1) \times (-1) + 1 \times a & (-1) \times 1 + 1 \times b \\a \times (-1) + b \times a & a \times 1 + b \times b \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Simplifying each element, we have:
\begin{bmatrix} 1 + a & -1 + b \\ -a + ab & a + b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}By comparing corresponding elements of the matrices, we get:
1 + a = 1
∴ a = 0.-1 + b = 0
∴ b = 1.
Therefore, the values are a = 0 and b = 1.
Question 12(i)
Solve for x and y :
\begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}We are given the matrix equation:
\begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}This implies:
\begin{bmatrix} 2 \times x + 5 \times y \\ 5 \times x + 2 \times y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}Thus, we have two equations:
- 2x + 5y = -7
- 5x + 2y = 14
Let’s solve these equations. Starting with the first equation:
2x + 5y = -7Rearranging, we find:
2x = -(7 + 5y)
x = \dfrac{-(7 + 5y)}{2} \quad \text{...(i)}
Next, substitute this expression for x into the second equation:
5x + 2y = 14Substitute x from equation (i):
5 \times \dfrac{-(7 + 5y)}{2} + 2y = 14Simplifying gives:
\dfrac{-35 - 25y}{2} + 2y = 14Combine terms:
\dfrac{-35 - 25y + 4y}{2} = 14Multiply through by 2 to clear the fraction:
-35 - 21y = 28Rearrange to solve for y:
-21y = 28 + 35
-21y = 63
y = -3
Now, substitute y = -3 back into equation (i) to find x:
x = \dfrac{-(7 + 5(-3))}{2}Simplify:
x = \dfrac{-(7 - 15)}{2}
x = \dfrac{8}{2}
x = 4
Therefore, the solution is x = 4 and y = -3.
Question 12(ii)
Solve for x and y :
\begin{bmatrix} x + y & x - 4 \end{bmatrix}\begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}We start with the given matrix equation:
\begin{bmatrix} x + y & x - 4 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}Let’s perform the matrix multiplication:
\Rightarrow \begin{bmatrix} (x + y) \times -1 + (x - 4) \times 2 & (x + y) \times -2 + (x - 4) \times 2 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}Simplifying the expression, we have:
\Rightarrow \begin{bmatrix} -x - y + 2x - 8 & -2x - 2y + 2x - 8 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}This reduces to:
\Rightarrow \begin{bmatrix} x - y - 8 & - 2y - 8 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}Now, equating the corresponding elements of the matrices, we find:
For the second element:
-2y - 8 = -11Solving for y:
\Rightarrow -2y = -3 \Rightarrow y = \dfrac{3}{2}For the first element:
x - y - 8 = -7Substituting the value of y:
\Rightarrow x - y = 1 \Rightarrow x = 1 + y \Rightarrow x = 1 + \dfrac{3}{2} = \dfrac{5}{2}Thus, the values are x = \dfrac{5}{2} and y = \dfrac{3}{2}.
Question 12(iii)
Solve for x and y :
\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3\begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2\begin{bmatrix} y \\ 3 \end{bmatrix}.
Let’s consider the given equation:
\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix}\begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3\begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2\begin{bmatrix} y \\ 3 \end{bmatrix}Perform the matrix multiplication and addition:
- The first matrix multiplication results in:
\begin{bmatrix} (-2) \times (-1) + 0 \times 2x \\ 3 \times (-1) + 1 \times 2x \end{bmatrix} = \begin{bmatrix} 2 \\ -3 + 2x \end{bmatrix} - Adding this to the matrix 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} gives:
\begin{bmatrix} 2 \\ -3 + 2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 + (-6) \\ -3 + 2x + 3 \end{bmatrix} = \begin{bmatrix} -4 \\ 2x \end{bmatrix}
This must equal the matrix on the right side:
\begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}Now, equate corresponding elements:
- From the first row: 2y = -4
- Solving gives y = -2.
- From the second row: 2x = 6
- Solving gives x = 3.
Therefore, the values are x = 3 and y = -2.
Question 13
In each case given below, find :
(a) the order of matrix M.
(b) the matrix M.
(i) M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} =\begin{bmatrix} 1 & 2 \end{bmatrix}
(ii) \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} \times M = \begin{bmatrix} 13 \\ 5 \end{bmatrix}
(i) Assume matrix M has dimensions a \times b.
The expression is given as:
M_{a \times b} \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}_{2 \times 2} = \begin{bmatrix} 1 & 2 \end{bmatrix}_{1 \times 2}Since matrix multiplication is possible only when the number of columns in the first matrix matches the number of rows in the second, we have b = 2.
Also, the number of rows in the resulting matrix equals the number of rows in the first matrix, so a = 1.
Thus, the order of matrix M is 1 \times 2.
Let M = \begin{bmatrix} x & y \end{bmatrix}.
We have:
\Rightarrow \begin{bmatrix} x & y \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \Rightarrow \begin{bmatrix} x \times 1 + y \times 0 & x \times 1 + y \times 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \Rightarrow \begin{bmatrix} x & x + 2y \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}From matrix equality, we deduce:
x = 1 x + 2y = 2 ⇒ 1 + 2y = 2 ⇒ 2y = 1⇒ y = \dfrac{1}{2}.
∴ M = \begin{bmatrix} x & y \end{bmatrix} = \begin{bmatrix} 1 & \dfrac{1}{2} \end{bmatrix}.
Thus, M = \begin{bmatrix} 1 & \dfrac{1}{2} \end{bmatrix}.
(ii) Suppose matrix M has dimensions a \times b.
The given equation is:
\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}_{2 \times 2} \times M_{a \times b} = \begin{bmatrix} 13 \\ 5 \end{bmatrix}_{2 \times 1}Matrix multiplication is valid only when the number of columns in the first matrix equals the number of rows in the second matrix, so a = 2.
Additionally, the number of columns in the resulting matrix matches the number of columns in the second matrix, hence b = 1.
Thus, the order of matrix M is 2 \times 1.
Let M = \begin{bmatrix} x \\ y \end{bmatrix}.
Then:
\Rightarrow \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 1 \times x + 4 \times y \\ 2 \times x + 1 \times y \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix} \Rightarrow \begin{bmatrix} x + 4y \\ 2x + y \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix}From matrix equality, we find:
x + 4y = 13⇒ x = 13 - 4y ……(i)
2x + y = 5Substituting (i) in the second equation:
⇒ 2(13 - 4y) + y = 5 ⇒ 26 - 8y + y = 5 ⇒ -7y = -21⇒ y = 3.
⇒ x = 13 - 4y = 13 - 4(3) = 13 - 12 = 1.
∴ M = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}.
Hence, M = \begin{bmatrix} 1 \\ 3 \end{bmatrix}.
Question 14
If A = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \text{and } B = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}; find the value of x, given that : A^2 = B.
We know that A^2 = B. This means:
\begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}Carrying out the matrix multiplication on the left side, we compute:
- For the first row, first column: (2 \times 2) + (x \times 0) = 4 + 0 = 4.
- For the first row, second column: (2 \times x) + (x \times 1) = 2x + x = 3x.
- For the second row, first column: (0 \times 2) + (1 \times 0) = 0 + 0 = 0.
- For the second row, second column: (0 \times x) + (1 \times 1) = 0 + 1 = 1.
Thus, the product of the matrices is:
\begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix}Equating this result with matrix B, we have:
\begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}From the equality of matrices, notice that 3x = 36. Solving for x gives:
x = \frac{36}{3} = 12Hence, x = 12.
Question 15
If A and B are any two 2 × 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?
Given that AB = BA = B and matrix B is not the zero matrix, we can deduce something important about matrix A.
This implies that A acts as a unit or identity matrix.
Question 16
Given A = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}, B = \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} and that AB = A + B; find the values of a, b and c.
We start with the equation:
AB = A + BSubstituting the given matrices, we have:
\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} a & b \\ 0 & c \end{bmatrix}Performing matrix multiplication on the left side, we get:
\begin{bmatrix} 3 \times a + 0 \times 0 & 3 \times b + 0 \times c \\ 0 \times a + 4 \times 0 & 0 \times b + 4 \times c \end{bmatrix} = \begin{bmatrix} 3 + a & b \\ 0 & 4 + c \end{bmatrix}This simplifies to:
\begin{bmatrix} 3a & 3b \\ 0 & 4c \end{bmatrix} = \begin{bmatrix} 3 + a & b \\ 0 & 4 + c \end{bmatrix}Using the equality of matrices, we equate corresponding elements:
For the first element, 3a = 3 + a. Solving this gives:
3a - a = 3
2a = 3
a = \dfrac{3}{2}For the second element, 3b = b. Solving this gives:
3b - b = 0
2b = 0
b = 0For the fourth element, 4c = 4 + c. Solving this gives:
4c - c = 4
3c = 4
c = \dfrac{4}{3}
Hence, a = \dfrac{3}{2}, b = 0, and c = \dfrac{4}{3}.
Question 17
If A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \text{ and } B = \begin{bmatrix} 3 \\ -11 \end{bmatrix} find the matrix X such that AX = B.
We have the equation:
AX = BThis translates to:
\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}X = \begin{bmatrix} 3 \\ -11 \end{bmatrix}The matrix X is of order 2 × 1, so let’s assume X = \begin{bmatrix} a \\ b \end{bmatrix}.
Substituting, we have:
\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 3 \\ -11 \end{bmatrix}On performing matrix multiplication, we get:
\begin{bmatrix} 2 \times a + 1 \times b \\ 1 \times a + 3 \times b \end{bmatrix} = \begin{bmatrix} 3 \\ -11 \end{bmatrix}This simplifies to:
\begin{bmatrix} 2a + b \\ a + 3b \end{bmatrix} = \begin{bmatrix} 3 \\ -11 \end{bmatrix}From the equality of matrices, we derive two equations:
- 2a + b = 3
Solving for b, we have:
b = 3 - 2a
( \text{… (i)} )
- a + 3b = -11
Substitute the value of b from equation (i) into the second equation:
a + 3(3 - 2a) = -11Simplifying, we find:
a + 9 - 6a = -11
-5a = -11 - 9
-5a = -20
a = 4
Now, substitute a = 4 back into equation (i):
b = 3 - 2(4)
b = 3 - 8
b = -5
Thus, the matrix X is:
X = \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \end{bmatrix}Question 18
If M = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}, show that : 6M – M^2 = 9I; where I is a 2 × 2 unit matrix.
To find M^2, we multiply the matrix M by itself:
M^2 = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}Calculating each element of the resulting matrix:
- First row, first column: 4 \times 4 + 1 \times (-1) = 16 - 1 = 15.
- First row, second column: 4 \times 1 + 1 \times 2 = 4 + 2 = 6.
- Second row, first column: -1 \times 4 + 2 \times (-1) = -4 - 2 = -6.
- Second row, second column: -1 \times 1 + 2 \times 2 = -1 + 4 = 3.
Thus, M^2 = \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix}.
Next, substitute M^2 into the expression 6M - M^2:
6\begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix}Calculate 6M:
6M = \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix}Now perform the subtraction:
- First row, first column: 24 - 15 = 9.
- First row, second column: 6 - 6 = 0.
- Second row, first column: -6 - (-6) = 0.
- Second row, second column: 12 - 3 = 9.
The result is \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}.
This can be expressed as 9 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, which is 9I, the identity matrix multiplied by 9.
∴ L.H.S. = R.H.S., confirming that 6M - M^2 = 9I.
Hence, proved that 6M – M^2 = 9I.
Question 19
If P = \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix} \text{ and Q} = \begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix}, find x and y such that PQ = null matrix.
We start by noting that the product of matrices P and Q results in a null matrix.
Thus, we have:
\begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix}\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}Performing matrix multiplication, we get:
\Rightarrow \begin{bmatrix} 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \\ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}Simplifying, the matrix becomes:
\Rightarrow \begin{bmatrix} 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}According to matrix equality, each corresponding element must be equal. Therefore:
- 6 + 6y = 0
- Solving for y, we get 6y = -6
Dividing both sides by 6, y = -1.
- 2x + 12 = 0
- Solving for x, we have 2x = -12
- Dividing both sides by 2, x = -6.
Hence, x = -6 and y = -1.
Question 20
Evaluate :
\begin{bmatrix} 2cos60° & -2sin30° \\ -tan 45° & cos0° \end{bmatrix}\begin{bmatrix} cot 45° & cosec 30° \\ sec60° & sin 90° \end{bmatrix}We start by evaluating the trigonometric expressions in the matrices:
For the first matrix:
– 2\cos60° = 2 \times \dfrac{1}{2} = 1
– -2\sin30° = -2 \times \dfrac{1}{2} = -1
– -\tan 45° = -1
– \cos0° = 1
The first matrix simplifies to:
\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}
For the second matrix:
– \cot 45° = 1
– \cosec 30° = 2
– \sec60° = 2
– \sin 90° = 1
The second matrix becomes:
\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}
Now, multiply the two matrices:
\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}
Perform the matrix multiplication:
– First row, first column: 1 \times 1 + (-1) \times 2 = 1 - 2 = -1
– First row, second column: 1 \times 2 + (-1) \times 1 = 2 - 1 = 1
– Second row, first column: -1 \times 1 + 1 \times 2 = -1 + 2 = 1
– Second row, second column: -1 \times 2 + 1 \times 1 = -2 + 1 = -1
Thus, the resulting matrix is:
\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}
Therefore, the product of the given matrices is:
\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}
Question 21
State, with reason, whether the following are true or false. A, B and C are matrices of order 2 × 2.
(i) A + B = B + A
(ii) A – B = B – A
(iii) (B.C).A = B.(C.A)
(iv) (A + B).C = A.C + B.C
(v) A.(B – C) = A.B – A.C
(vi) (A – B).C = A.C – B.C
(vii) A^2 – B^2 = (A + B)(A – B)
(viii) (A – B)^2 = A^2 – 2A.B + B^2.
(i) A + B = B + A
This statement holds true since matrix addition is commutative, meaning the order of addition does not affect the result.
(ii) A – B = B – A
This statement is false because matrix subtraction lacks commutativity; swapping the matrices changes the result.
(iii) (B.C).A = B.(C.A)
This statement is true as matrix multiplication is associative, allowing us to regroup the matrices without altering the product.
(iv) (A + B).C = A.C + B.C
This statement is true because matrix multiplication distributes over addition, much like in regular arithmetic.
(v) A.(B – C) = A.B – A.C
This statement is true since matrix multiplication is distributive over subtraction, allowing us to multiply each term separately.
(vi) (A – B).C = A.C – B.C
This statement is true as matrix multiplication distributes over subtraction, similar to how it does over addition.
(vii) A^2 – B^2 = (A + B)(A – B)
This statement is false because matrix operations do not adhere to the same algebraic factorization and expansion rules as numbers.
(viii) (A – B)^2 = A^2 – 2A.B + B^2
This statement is false because the typical algebraic expansion rules are not applicable to matrices.
Test Yourself
Question 1(a)
If a matrix A = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} and matrix B = \begin{bmatrix} 3 \\ 1 \end{bmatrix}, then which of the following is possible :
- (a) A + B
- (b) A – B
- (c) AB
- (d) BA
For matrix addition or subtraction to occur, both matrices must share the same dimensions. Since matrix A is 2 \times 2 and matrix B is 2 \times 1, operations like A + B and A – B cannot be performed.
Now, consider matrix multiplication. The product AB is feasible because the number of columns in matrix A (which is 2) matches the number of rows in matrix B (also 2).
However, the multiplication BA is not feasible. This is due to matrix B having 1 column, which does not equal the 2 rows of matrix A.
Hence, Option 3 is the correct option.
Question 1(b)
If M × \begin{bmatrix} 3 & 2 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \end{bmatrix}, the order of matrix M is :
- (a) 2 × 2
- (b) 2 × 1
- (c) 1 × 2
- (d) 1 × 3
To determine the order of matrix M, consider the rules of matrix multiplication. The number of columns in the first matrix must match the number of rows in the second matrix. Additionally, the resulting matrix’s dimensions are defined by the number of rows from the first matrix and the number of columns from the second matrix.
Assume the order of matrix M is a \times b. We have:
M_{a \times b} \times \begin{bmatrix} 3 & 2 \\ -1 & 0 \end{bmatrix}_{2 \times 2} = \begin{bmatrix} 3 & -1 \end{bmatrix}_{1 \times 2}Since the second matrix has 2 rows, matrix M must have 2 columns, so b = 2. The resultant matrix has 1 row, indicating that matrix M must have 1 row, so a = 1.
Thus, the order of matrix M is 1 \times 2.
Hence, Option 3 is the correct option.
Question 1(c)
If \begin{bmatrix} 2x - y \\ x + y \end{bmatrix} = \begin{bmatrix} 9 \\ 9 \end{bmatrix}, the value of x and y are :
- (a) x = 3 and y = 3
- (b) x = 3 and y = 9
- (c) x = 3 and y = 6
- (d) x = 6 and y = 3
We start by examining the given matrix equation:
\begin{bmatrix} 2x - y \\ x + y \end{bmatrix} = \begin{bmatrix} 9 \\ 9 \end{bmatrix}This implies two equations:
- 2x - y = 9
- x + y = 9
Let’s add these two equations to eliminate y:
2x - y + x + y = 9 + 9This simplifies to:
3x = 18Solving for x, we divide both sides by 3:
x = \dfrac{18}{3} = 6Now, substitute x = 6 back into equation (2):
6 + y = 9Solving for y, we get:
y = 9 - 6 = 3Thus, the values are x = 6 and y = 3. Option 4 is the correct option.
Question 1(d)
If matrix A = \begin{bmatrix} x - y & x + y \\ y - x & y + x \end{bmatrix} and matrix B = \begin{bmatrix} x + y & y - x \\ x - y & y + x \end{bmatrix}, then A + B is :
- (a) \begin{bmatrix} 2y & 2x \\ 0 & 2(x + y) \end{bmatrix}
- (b) \begin{bmatrix} 2x & 2(x + y) \\ 0 & 0 \end{bmatrix}
- (c) \begin{bmatrix} 2x & 2y \\ 0 & 2(x + y) \end{bmatrix}
- (d) \begin{bmatrix} 2x - 2y & 2y \\ 0 & 0 \end{bmatrix}
To find the sum of matrices A and B, we substitute their values into the expression A + B:
A + B = \begin{bmatrix} x - y & x + y \\ y - x & y + x \end{bmatrix} + \begin{bmatrix} x + y & y - x \\ x - y & y + x \end{bmatrix}When we add these matrices, we combine the corresponding elements:
= \begin{bmatrix} (x - y) + (x + y) & (x + y) + (y - x) \\ (y - x) + (x - y) & (y + x) + (y + x) \end{bmatrix}Simplifying each element, notice that:
– The first element becomes x - y + x + y = 2x.
– The second element becomes x + y + y - x = 2y.
– The third element simplifies to y - x + x - y = 0.
– The fourth element becomes ( y + x + y + x = 2(x + y) ).
Thus, the resulting matrix is:
\begin{bmatrix} 2x & 2y \\ 0 & 2(x + y) \end{bmatrix}Hence, Option 3 is the correct option.
Question 1(e)
Event A : Order of matrix A is 3 × 5.
Event B : Order of matrix B is 5 × 3.
Event C : Order of matrix C is 3 × 3.
Product of which two matrices gives a square matrix.
- (a) AB and AC
- (b) AB and BC
- (c) BA and BC
- (d) AB and BA
To determine if the product of two matrices is a square matrix, we must consider the dimensions of the matrices involved. When multiplying two matrices, the resulting matrix has an order determined by the number of rows in the first matrix and the number of columns in the second matrix.
For the product AB, the order is 3 × 3 because matrix A has 3 rows and matrix B has 3 columns. This gives us a square matrix of order 3.
For the product BA, the order is 5 × 5 because matrix B has 5 rows and matrix A has 5 columns. This results in a square matrix of order 5.
Thus, both products AB and BA result in square matrices.
Hence, Option 4 is the correct option.
Question 1(f)
Two matrices A and B each of order 2 x 2.
Assertion (A) : A X B = 0 \nRightarrow A = 0 or B = 0.
Reason (R) : Let A = \begin{bmatrix} 2 & 2 \\ 5 & 5 \end{bmatrix} ≠ 0 and B = \begin{bmatrix} -4 & 3 \\ 4 & -3 \end{bmatrix} ≠ 0 but A x B = \begin{bmatrix} 2 & 2 \\ 5 & 5 \end{bmatrix}\begin{bmatrix} -4 & 3 \\ 4 & -3 \end{bmatrix} = 0.
- (a) A is true, R is false.
- (b) A is false, R is true.
- (c) Both A and R are true and R is correct reason for A.
- (d) Both A and R are true and R is incorrect reason for A.
In matrix algebra, it is not always the case that if the product of two matrices A \times B = 0, then either A = 0 or B = 0. This makes the assertion (A) true.
Consider the given matrices A = \begin{bmatrix} 2 & 2 \\ 5 & 5 \end{bmatrix} and B = \begin{bmatrix} -4 & 3 \\ 4 & -3 \end{bmatrix}. Both matrices are non-zero.
Now, compute the product AB:
AB = \begin{bmatrix} 2 & 2 \\ 5 & 5 \end{bmatrix} \cdot \begin{bmatrix} -4 & 3 \\ 4 & -3 \end{bmatrix} = \begin{bmatrix} 2 \times (-4) + 2 \times 4 & 2 \times 3 + 2 \times (-3) \\5 \times (-4) + 5 \times 4 & 5 \times 3 + 5 \times (-3) \end{bmatrix}Simplifying the calculations:
= \begin{bmatrix} -8 + 8 & 6 - 6 \\-20 + 20 & 15 - 15 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\0 & 0 \end{bmatrix}This shows that the product AB = 0, even though neither A nor B is a zero matrix. Thus, the reason (R) is also true.
Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion. Option 3 is the correct option.
Question 1(g)
Matrix A = \begin{bmatrix} x & y \end{bmatrix} and Matrix B = \begin{bmatrix} a \\ b \end{bmatrix}.
Assertion (A) : Product BA is possible and order of resulting matrix is 2 x 2.
Reason (R) : The product BA of two matrices A and B is possible only if number of rows in matrix B Is same as number of columns in matrix A.
- (a) A is true, R is false.
- (b) A is false, R is true.
- (c) Both A and R are true and R is correct reason for A.
- (d) Both A and R are true and R is incorrect reason for A.
Matrix A has dimensions 1 x 2, while Matrix B is 2 x 1. For the multiplication of matrices to be feasible, the number of columns in the first matrix must match the number of rows in the second matrix.
Here, the number of columns in matrix B is 1, and the number of rows in matrix A is 1, which satisfies the requirement for multiplying them in the order BA.
The resulting matrix from multiplying BA will have the number of rows equal to matrix B and the number of columns equal to matrix A. Therefore, the order of matrix BA is 2 x 2.
Thus, the assertion is indeed correct. However, the reason provided is incorrect because it inaccurately states the condition for matrix multiplication. The correct condition is that the number of columns in the first matrix must match the number of rows in the second matrix, not the other way around.
Hence, option 1 is the correct option.
Question 1(h)
A, B and C are three matrices each of order 2 x 2.
Statement 1 : If A x B = A x C ⇒ B = C
Statement 2 : Cancellation law is applicable in matrix multiplication.
- (a) Both the statements are true.
- (b) Both the statements are false.
- (c) Statement 1 is true, and statement 2 is false.
- (d) Statement 1 is false, and statement 2 is true.
Matrix multiplication does not adhere to the cancellation law. This means that even if you have A x B = A x C, you cannot deduce that B equals C. Therefore, Statement 2 is incorrect. Consequently, Statement 1 is also incorrect because it relies on the cancellation law being valid.
Hence, option 2 is the correct option.
Question 1(i)
Matrix A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} and matrix B = \begin{bmatrix} 5 & 5 \\ 5 & 5 \end{bmatrix}
Statement 1 : AB = 0
Statement 2 : AB = 0, even if A ≠ 0 and B ≠ 0.
- (a) Both the statements are true.
- (b) Both the statements are false.
- (c) Statement 1 is true, and statement 2 is false.
- (d) Statement 1 is false, and statement 2 is true.
Consider the matrices given:
Matrix A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} and Matrix B = \begin{bmatrix} 5 & 5 \\ 5 & 5 \end{bmatrix}.
Let’s calculate the product AB:
\Rightarrow AB = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \cdot \begin{bmatrix} 5 & 5 \\ 5 & 5 \end{bmatrix}This results in:
= \begin{bmatrix} 2 \times 5 + (-2) \times 5 & 2 \times 5 + (-2) \times 5 \\ (-2) \times 5 + 2 \times 5 & (-2) \times 5 + 2 \times 5 \end{bmatrix}Simplifying the calculations, we find:
= \begin{bmatrix} 10 - 10 & 10 - 10 \\ -10 + 10 & -10 + 10 \end{bmatrix}Thus, we arrive at:
= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}This shows that AB = 0. Notice that both A and B are non-zero matrices, yet their product is the zero matrix. Therefore, both statements are indeed true.
Hence, option 1 is the correct option.
Question 2
Find x and y, if :
\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}\begin{bmatrix} 2x \\ 1 \end{bmatrix} + 2\begin{bmatrix} -4 \\ 5 \end{bmatrix} = 4\begin{bmatrix} 2 \\ y \end{bmatrix}We’re given the equation:
\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}\begin{bmatrix} 2x \\ 1 \end{bmatrix} + 2\begin{bmatrix} -4 \\ 5 \end{bmatrix} = 4\begin{bmatrix} 2 \\ y \end{bmatrix}First, perform the matrix multiplication:
\begin{bmatrix} 3 \times 2x + (-2) \times 1 \\ -1 \times 2x + 4 \times 1 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix} = \begin{bmatrix} 8 \\ 4y \end{bmatrix}This results in:
\begin{bmatrix} 6x - 2 \\ -2x + 4 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix} = \begin{bmatrix} 8 \\ 4y \end{bmatrix}Next, add the matrices on the left side:
\begin{bmatrix} 6x - 2 + (-8) \\ -2x + 4 + 10 \end{bmatrix} = \begin{bmatrix} 8 \\ 4y \end{bmatrix}This simplifies to:
\begin{bmatrix} 6x - 10 \\ -2x + 14 \end{bmatrix} = \begin{bmatrix} 8 \\ 4y \end{bmatrix}According to matrix equality, equate the corresponding elements:
For the first element:
6x - 10 = 8⇒ 6x = 18
⇒ x = 3
For the second element:
-2x + 14 = 4ySubstitute x = 3:
-2(3) + 14 = 4y⇒ -6 + 14 = 4y
⇒ 8 = 4y
⇒ y = 2
Thus, x = 3 and y = 2.
Question 3
Find x and y, if :
\begin{bmatrix} 3x & 8 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} - 3\begin{bmatrix} 2 & -7 \end{bmatrix} = 5\begin{bmatrix} 3 & 2y \end{bmatrix}We are given the equation:
\begin{bmatrix} 3x & 8 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} - 3\begin{bmatrix} 2 & -7 \end{bmatrix} = 5\begin{bmatrix} 3 & 2y \end{bmatrix}Let’s solve it step by step. First, perform the matrix multiplication on the left:
\Rightarrow \begin{bmatrix} 3x \times 1 + 8 \times 3 & 3x \times 4 + 8 \times 7 \end{bmatrix}This simplifies to:
\Rightarrow \begin{bmatrix} 3x + 24 & 12x + 56 \end{bmatrix}Subtract the matrix \begin{bmatrix} 6 & -21 \end{bmatrix} from this result:
\Rightarrow \begin{bmatrix} 3x + 24 - 6 & 12x + 56 - (-21) \end{bmatrix}Simplifying further, we get:
\Rightarrow \begin{bmatrix} 3x + 18 & 12x + 77 \end{bmatrix}This must equal the matrix \begin{bmatrix} 15 & 10y \end{bmatrix}.
By comparing corresponding elements, we have the equations:
- 3x + 18 = 15
- 12x + 77 = 10y
Solving the first equation:
3x + 18 = 15
\Rightarrow 3x = -3
\Rightarrow x = -1
Now, solve the second equation with x = -1:
12(-1) + 77 = 10y
\Rightarrow -12 + 77 = 10y
\Rightarrow 65 = 10y
\Rightarrow y = 6.5
Hence, x = -1 and y = 6.5
Question 4
If \begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix} \text{ and } \begin{bmatrix} -x & y \end{bmatrix}\begin{bmatrix} 2x \\ y \end{bmatrix} = \begin{bmatrix} -2 \end{bmatrix};
find x and y, if :
(i) x, y ∈ W (whole numbers)
(ii) x, y ∈ Z (integers)
We start with the given matrix equation:
\begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} 25 \end{bmatrix}Performing the matrix multiplication, we get:
\begin{bmatrix} x \times x + y \times y \end{bmatrix} =\begin{bmatrix} 25 \end{bmatrix}This simplifies to:
\begin{bmatrix} x^2 + y^2 \end{bmatrix} =\begin{bmatrix} 25 \end{bmatrix}By comparing the elements, we get our first equation:
x^2 + y^2 = 25This can be written as:
x^2 = 25 - y^2 \quad \text{...(i)}Next, consider the second matrix equation:
\begin{bmatrix} -x & y \end{bmatrix}\begin{bmatrix} 2x \\ y \end{bmatrix} =\begin{bmatrix} -2 \end{bmatrix}Again, multiplying the matrices gives:
\begin{bmatrix} -x \times 2x + y \times y \end{bmatrix} =\begin{bmatrix} -2 \end{bmatrix}This results in:
\begin{bmatrix} -2x^2 + y^2 \end{bmatrix} =\begin{bmatrix} -2 \end{bmatrix}By comparing the elements, we get our second equation:
-2x^2 + y^2 = -2 \quad \text{...(ii)}Substitute x^2 from equation (i) into equation (ii):
-2(25 - y^2) + y^2 = -2Simplifying, we have:
-50 + 2y^2 + y^2 = -2 3y^2 = -2 + 50 3y^2 = 48 y^2 = 16Thus, y = \pm 4.
Substitute back to find x:
x^2 = 25 - y^2 x^2 = 25 - 16 x^2 = 9x = \pm 3.
(i) For whole numbers x, y \in W:
∴ x = 3, y = 4.
Hence, x = 3 and y = 4.
(ii) For integers x, y \in Z:
∴ x = \pm 3, y = \pm 4.
Hence, x = \pm 3 and y = \pm 4.
Question 5
Evaluate :
\begin{bmatrix} cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix}\begin{bmatrix} sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix}We start with the given matrices:
\begin{bmatrix} \cos 45° & \sin 30° \\ \sqrt{2}\cos 0° & \sin 0° \end{bmatrix} \begin{bmatrix} \sin 45° & \cos 90° \\ \sin 90° & \cot 45° \end{bmatrix}Substituting the trigonometric values, we have:
\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} \\ \sqrt{2}(1) & 0 \end{bmatrix} \begin{bmatrix} \dfrac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix}Simplifying further, this becomes:
\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} \\ \sqrt{2} & 0 \end{bmatrix} \begin{bmatrix} \dfrac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix}Now, perform the matrix multiplication:
\begin{bmatrix} \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \times 1 & \dfrac{1}{\sqrt{2}} \times 0 + \dfrac{1}{2} \times 1 \\ \sqrt{2} \times \dfrac{1}{\sqrt{2}} + 0 \times 1 & \sqrt{2} \times 0 + 0 \times 1 \end{bmatrix}Calculating each element, we get:
\begin{bmatrix} \dfrac{1}{2} + \dfrac{1}{2} & 0 + \dfrac{1}{2} \\ 1 + 0 & 0 + 0 \end{bmatrix}This simplifies to:
\begin{bmatrix} 1 & \dfrac{1}{2} \\ 1 & 0 \end{bmatrix}Finally, in decimal form:
\begin{bmatrix} 1 & 0.5 \\ 1 & 0 \end{bmatrix}Thus, the product of the given matrices is \begin{bmatrix} 1 & 0.5 \\ 1 & 0 \end{bmatrix}.
Question 6
If A = \begin{bmatrix} 0 & -1 \\ 4 & -3 \end{bmatrix}, B = \begin{bmatrix} -5 \\ 6 \end{bmatrix} and 3A × M = 2B; find matrix M.
Consider the order of matrix M as a \times b.
We have the equation:
3A \times M = 2B \tag{i}This implies:
3\begin{bmatrix} 0 & -1 \\ 4 & -3 \end{bmatrix}_{2 \times 2} \times M_{a \times b} = 2\begin{bmatrix} -5 \\ 6 \end{bmatrix}_{2 \times 1}For matrix multiplication to be valid, the number of columns in the first matrix must match the number of rows in the second matrix.
∴ a = 2.
Furthermore, the number of columns in the resulting matrix equals the number of columns in the second matrix.
∴ b = 1.
Thus, the order of matrix M is 2 \times 1.
Assume:
M = \begin{bmatrix} a \\ b \end{bmatrix}Substitute the values of A, M, and B into equation (i):
3\begin{bmatrix} 0 & -1 \\ 4 & -3 \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = 2\begin{bmatrix} -5 \\ 6 \end{bmatrix}This simplifies to:
\begin{bmatrix} 0 & -3 \\ 12 & -9 \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} -10 \\ 12 \end{bmatrix}Breaking it down further:
\begin{bmatrix} 0 \times a + (-3) \times b \\ 12 \times a + (-9) \times b \end{bmatrix} = \begin{bmatrix} -10 \\ 12 \end{bmatrix}Thus, we have:
\begin{bmatrix} -3b \\ 12a - 9b \end{bmatrix} = \begin{bmatrix} -10 \\ 12 \end{bmatrix}From the equality of matrices, we derive:
-3b = -10⇒ b = \dfrac{10}{3} ( \text{…(i)} )
And:
12a - 9b = 12Substituting the value of b from equation (i):
12a - 9 \times \dfrac{10}{3} = 12 12a - 30 = 12 12a = 42 a = \dfrac{7}{2}∴ The matrix M is:
\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix}Therefore, matrix M is:
M = \begin{bmatrix} \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix}Question 7
Find x and y if : \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}.
We start with the matrix equation:
\begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}
Performing matrix multiplication, we calculate:
\begin{bmatrix} x \times 2 + 3x \times 1 \\ y \times 2 + 4y \times 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}
This simplifies to:
\begin{bmatrix} 5x \\ 6y \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}
Comparing the corresponding elements of the matrices, we have:
For the first row,
5x = 5
⇒ x = 1.
For the second row,
6y = 12
⇒ y = 2.
Hence, x = 1 and y = 2.
Question 8
If matrix X = \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \end{bmatrix} \text{ and 2X - 3Y} = \begin{bmatrix} 10 \\ -8 \end{bmatrix}, find the matrix ‘X’ and matrix ‘Y’.
Let’s evaluate the matrix X first. We have:
X = \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \end{bmatrix}To find the product, calculate each element of the resulting matrix:
X = \begin{bmatrix} (-3) \times 2 + 4 \times (-2) \\ 2 \times 2 + (-3) \times (-2) \end{bmatrix}Simplifying, we get:
X = \begin{bmatrix} -6 + (-8) \\ 4 + 6 \end{bmatrix} X = \begin{bmatrix} -14 \\ 10 \end{bmatrix}Now, using the equation 2X - 3Y = \begin{bmatrix} 10 \\ -8 \end{bmatrix}, substitute X:
2\begin{bmatrix} -14 \\ 10 \end{bmatrix} - 3Y = \begin{bmatrix} 10 \\ -8 \end{bmatrix}Multiply the matrix X by 2:
\begin{bmatrix} -28 \\ 20 \end{bmatrix} - 3Y = \begin{bmatrix} 10 \\ -8 \end{bmatrix}Now, isolate 3Y:
3Y = \begin{bmatrix} -28 \\ 20 \end{bmatrix} - \begin{bmatrix} 10 \\ -8 \end{bmatrix}Subtract the matrices:
3Y = \begin{bmatrix} -28 - 10 \\ 20 - (-8) \end{bmatrix} 3Y = \begin{bmatrix} -38 \\ 28 \end{bmatrix}Finally, divide by 3 to find Y:
Y = \dfrac{1}{3}\begin{bmatrix} -38 \\ 28 \end{bmatrix}Therefore, X = \begin{bmatrix} -14 \\ 10 \end{bmatrix} and Y = \dfrac{1}{3}\begin{bmatrix} -38 \\ 28 \end{bmatrix}.”}
Question 9
If A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}, B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} and I is the identity matric of same order and A^t is transpose of matrix A, find A^t.B + BI.
Given matrices are A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} and B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}. The transpose of matrix A, denoted as A^t, is \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}. The identity matrix I of the same order is \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.
Now, let’s compute A^t.B + BI:
\Rightarrow A^t.B + BI = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Perform the matrix multiplication for A^t.B:
= \begin{bmatrix} 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix} = \begin{bmatrix} 8 - 1 & -4 + 3 \\ 20 - 3 & -10 + 9 \end{bmatrix} = \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix}Next, compute BI:
= \begin{bmatrix} 4 \times 1 + (-2) \times 0 & 4 \times 0 + (-2) \times 1 \\ -1 \times 1 + 3 \times 0 & -1 \times 0 + 3 \times 1 \end{bmatrix} = \begin{bmatrix} 4 + 0 & 0 - 2 \\ -1 + 0 & 0 + 3 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}Add the results of A^t.B and BI:
= \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 7 + 4 & -1 + (-2) \\ 17 + (-1) & -1 + 3 \end{bmatrix}= \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}.
∴ A^t.B + BI = \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}.
Question 10
Let A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}. Find A^2 + AB + B^2.
To find A^2, multiply matrix A by itself:
A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}.Next, calculate B^2 by multiplying matrix B by itself:
B^2 = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \\ -1 \times 2 + 0 \times (-1) & -1 \times 3 + 0 \times 0 \end{bmatrix} = \begin{bmatrix} 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix} = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}.Now, substitute the values of A^2 and B^2 into the expression A^2 + AB + B^2:
A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}Calculate AB:
AB = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \\ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix} = \begin{bmatrix} 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix}.Add the matrices:
A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix} = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}.∴ A^2 + AB + B^2 = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}.
Question 11
If A = \begin{bmatrix} 3 & a \\ -4 & 8 \end{bmatrix}, B = \begin{bmatrix} c & 4 \\ -3 & 0 \end{bmatrix}, C = \begin{bmatrix} -1 & 4 \\ 3 & b \end{bmatrix} and 3A – 2C = 6B, find the values of a, b and c.
We are given the equation:
3A - 2C = 6BSubstitute the matrices for A, B, and C:
3\begin{bmatrix} 3 & a \\ -4 & 8 \end{bmatrix} - 2\begin{bmatrix} -1 & 4 \\ 3 & b \end{bmatrix} = 6\begin{bmatrix} c & 4 \\ -3 & 0 \end{bmatrix}Calculate the scalar multiplication for each matrix:
\begin{bmatrix} 9 & 3a \\ -12 & 24 \end{bmatrix} - \begin{bmatrix} -2 & 8 \\ 6 & 2b \end{bmatrix} = \begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}Perform the subtraction of matrices:
\begin{bmatrix} 9 - (-2) & 3a - 8 \\ -12 - 6 & 24 - 2b \end{bmatrix} = \begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}This simplifies to:
\begin{bmatrix} 11 & 3a - 8 \\ -18 & 24 - 2b \end{bmatrix} = \begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}Since the matrices are equal, equate corresponding elements:
6c = 11
\Rightarrow c = \frac{11}{6} = 1\frac{5}{6}3a - 8 = 24
\Rightarrow 3a = 32
\Rightarrow a = \frac{32}{3} = 10\frac{2}{3}24 - 2b = 0
\Rightarrow 2b = 24
\Rightarrow b = 12
Thus, the values are: a = 10\frac{2}{3}, b = 12, c = 1\frac{5}{6}.
Question 12
Given A = \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}, B = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}, C = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} and BA = C^2. Find the values of p and q.
We start with the equation:
BA = C^2Substituting the given matrices, we have:
\begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix} \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}Now, calculate the product on the left-hand side:
\begin{bmatrix} 0 \times p + (-q) \times 0 & 0 \times 0 + (-q) \times 2 \\ 1 \times p + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix}Next, compute the product on the right-hand side:
\begin{bmatrix} 2 \times 2 + (-2) \times 2 & 2 \times (-2) + (-2) \times 2 \\ 2 \times 2 + 2 \times 2 & 2 \times (-2) + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix}By comparing the corresponding elements of the matrices, we have:
- From the first element of the second row, p = 8.
- From the second element of the first row, -2q = -8, which gives q = 4.
Hence, p = 8 and q = 4.
Question 13
Evaluate : \begin{bmatrix} 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix}\begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}.
To solve this matrix multiplication, we start by evaluating the trigonometric functions in the first matrix:
\Rightarrow \begin{bmatrix} 4 \sin 30° & 2\cos 60° \\ \sin 90° & 2 \cos 0° \end{bmatrix} = \begin{bmatrix} 4 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix}This simplifies to:
= \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}Next, we multiply this resulting matrix with the second matrix:
\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}We calculate the elements of the resulting matrix by performing the dot product of rows and columns:
= \begin{bmatrix} 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix}Simplifying each element, we get:
= \begin{bmatrix} 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix}Therefore, the final matrix is:
= \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}.Thus, the product of the given matrices is \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}.
Question 14
Given A = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} and A^2 = 9A + mI. Find m.
Start with the equation given:
A^2 = 9A + mISubstitute the given matrices for A and I:
\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} = 9 \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} + m \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Perform the matrix multiplication on the left side:
\begin{bmatrix} 2 \times 2 + 0 \times (-1) & 2 \times 0 + 0 \times 7 \\-1 \times 2 + 7 \times (-1) & -1 \times 0 + 7 \times 7 \end{bmatrix} = \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix} + \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}This results in:
\begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} = \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix} + \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}To isolate the matrix with m, subtract:
\begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} - \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix}Calculate the subtraction:
\begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} = \begin{bmatrix} 4 - 18 & 0 - 0 \\-9 - (-9) & 49 - 63 \end{bmatrix}This simplifies to:
\begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} = \begin{bmatrix} -14 & 0 \\ 0 & -14 \end{bmatrix}Since the matrices are equal, m must be the same in both positions:
∴ m = -14.
Hence, m = -14.
Question 15
Given matrix A = \begin{bmatrix} 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix}\text{ and } B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}. If AX = B,
(i) write the order of matrix X.
(ii) find the matrix ‘X’.
(i) Suppose the order of matrix X is a \times b.
This means we have:
\begin{bmatrix} 4 \sin 30° & \cos 0° \\ \cos 0° & 4 \sin 30° \end{bmatrix}_{2 \times 2} \times X_{a \times b} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}_{2 \times 1}Matrix multiplication is feasible only when the number of columns in the first matrix equals the number of rows in the second matrix.
∴ a = 2.
Also, the number of columns in the resulting matrix is the same as the number of columns in the second matrix.
∴ b = 1.
Hence, the order of matrix X is 2 \times 1.
(ii) Assume matrix X is \begin{bmatrix} x \\ y \end{bmatrix}.
Given that:
\Rightarrow AX = B \\\Rightarrow \begin{bmatrix} 4 \sin 30° & \cos 0° \\ \cos 0° & 4 \sin 30° \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \\\Rightarrow \begin{bmatrix} 4 \times \dfrac{1}{2} & 1 \\ 1 & 4 \times \dfrac{1}{2} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \\\Rightarrow \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \\\Rightarrow \begin{bmatrix} 2x + y \\ x + 2y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}From the equality of matrices, we have:
2x + y = 4⇒ y = 4 - 2x …….(i)
x + 2y = 5Substitute the expression for y from (i) into the second equation:
⇒ x + 2(4 - 2x) = 5
⇒ x + 8 - 4x = 5
⇒ -3x = 5 - 8
⇒ -3x = -3
⇒ x = 1.
Now, find y:
⇒ y = 4 - 2x = 4 - 2(1) = 2.
\therefore X = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}Hence, X = \begin{bmatrix} 1 \\ 2 \end{bmatrix}.
Case-Study Based Question
Question 1
Case study:
Neeta, Salma and Vivian are neighbours in different flats of the same building. Last Sunday, they went together to a departmental store to buy groceries.
The groceries purchased by them is as shown below.
| Pulses(kg) | Tea packets | Refined oil(litre) | |
|---|---|---|---|
| 1.Neeta | 15 | 3 | 8 |
| 2.Salma | 10 | 4 | 12 |
| 3.Vivian | 5 | 5 | 9 |
The prices of the items are as follows :
Pulses : ₹ 170 per kg
Tea : ₹ 150 per packet
Refined oil : ₹ 250 per litre
Using matrix multiplication, find the total amount that Neeta, Salma and Vivian paid.
To determine the total amount each person paid, we start by organizing the quantities of groceries purchased into a matrix.
The quantity matrix, Q, is as follows:
Q = \begin{bmatrix} 15 & 3 & 8 \\ 10 & 4 & 12 \\ 5 & 5 & 9 \end{bmatrix}Next, we consider the prices of the items, which we arrange in a price matrix, P:
P = \begin{bmatrix} 170 \\ 150 \\ 250 \end{bmatrix}To find the total expenditure for each person, we perform matrix multiplication of Q and P:
\Rightarrow QP = \begin{bmatrix} 15 & 3 & 8 \\ 10 & 4 & 12 \\ 5 & 5 & 9 \end{bmatrix} \times \begin{bmatrix} 170 \\ 150 \\ 250 \end{bmatrix}Calculating each entry of the resulting matrix:
\Rightarrow QP = \begin{bmatrix} 15 \times 170 + 3 \times 150 + 8 \times 250 \\ 10 \times 170 + 4 \times 150 + 12 \times 250 \\ 5 \times 170 + 5 \times 150 + 9 \times 250 \end{bmatrix}Breaking down the calculations:
\Rightarrow QP = \begin{bmatrix} 2550 + 450 + 2000 \\ 1700 + 600 + 3000 \\ 850 + 750 + 2250 \end{bmatrix}This simplifies to:
\Rightarrow QP = \begin{bmatrix} 5000 \\ 5300 \\ 3850 \end{bmatrix}∴ The amounts paid by Neeta, Salma, and Vivian are ₹ 5,000, ₹ 5,300, and ₹ 3,850 respectively.