Matrices — Multiple Choice Questions

If $\begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*},$ then the values of x and y are

1. x = 2, y = 7
2. x = 7, y = 2
3. x = 3, y = 6
4. x = -2, y = 7

Answer

Given,

$\begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*} \\[0.5em]$

By definition of equality of matrices we get,

⇒ x + 3 = 5 or x = 5 - 3 = 2.

⇒ y - 4 = 3 or y = 3 + 4 = 7.

⇒ x + y = 9.

Since, x = 2 and y = 7 satisfies the equation x + y = 9, hence, value of x = 2 and y = 7.

∴ Option 1 is the correct option.

If $\begin{bmatrix*}[r] x + 2y & -y \\ 3x & 4 \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 3 \\ 6 & 4 \end{bmatrix*},$ then the values of x and y are

1. x = 2, y = 3
2. x = 2, y = -3
3. x = -2, y = 3
4. x = 3, y = 2

Answer

Given,

$\begin{bmatrix*}[r] x + 2y & -y \\ 3x & 4 \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 3 \\ 6 & 4 \end{bmatrix*} \\[0.5em]$

By definition of equality of matrices we get,

⇒ x + 2y = -4     (...Eq 1)

⇒ -y = 3 or y = -3

⇒ 3x = 6 or x = 2

Putting, x = 2 and y = -3 in Eq 1,

⇒ x + 2y = -4
⇒ L.H.S. = 2 + 2(-3) = 2 - 6 = -4 = R.H.S.

Since, x = 2 and y = -3 satisfies the equation x + 2y = -4,

∴ x = 2 and y = -3.

∴ Option 2 is the correct option.

If $\begin{bmatrix*}[r] x - 2y & 5 \\ 3 & y \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 5 \\ 3 & -2 \end{bmatrix*},$ then the value of x is

1. -2
2. 0
3. 1
4. 2

Answer

Given,

$\begin{bmatrix*}[r] x - 2y & 5 \\ 3 & y \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 5 \\ 3 & -2 \end{bmatrix*} \\[0.5em]$

By definition of equality of matrices,

⇒ x - 2y = 6     (...Eq 1)

⇒ y = -2.

Putting value of y in Eq 1 we get,

⇒ x - 2y = 6
⇒ x - 2(-2) = 6
⇒ x + 4 = 6
⇒ x = 6 - 4
⇒ x = 2.

∴ x = 2.

∴ Option 4 is the correct option .

If $\begin{bmatrix*}[r] x + 2y & 3y \\ 4x & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ 8 & 2 \end{bmatrix*},$ then the value of x - y is

1. -3
2. 1
3. 3
4. 5

Answer

Given,

$\begin{bmatrix*}[r] x + 2y & 3y \\ 4x & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ 8 & 2 \end{bmatrix*} \\[0.5em]$

By definition of equality of matrices we get,

⇒ x + 2y = 0     (...Eq 1)

⇒ 3y = -3 or y = -1

⇒ 4x = 8 or x = 2

Putting the value of x = 2 and y = -1 in Eq 1

⇒ x + 2y = 0
⇒ L.H.S. = 2 + 2(-1) = 2 - 2 = 0 = R.H.S.

Since, x = 2 and y = -1 satisfies Eq 1

∴ x = 2, y = -1 and x - y = 2 - (-1) = 2 + 1 = 3.

∴ Option 3 is the correct option.

If $x\begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} + y\begin{bmatrix*}[r] -1 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*},$ then the values of x and y are

1. x = 2, y = 6
2. x = 2, y = -6
3. x = 3, y = -4
4. x = 3, y = -6

Answer

Given,

$x\begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} + y\begin{bmatrix*}[r] -1 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 3x \end{bmatrix*} + \begin{bmatrix*}[r] -y \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x - y \\ 3x \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em]$

By definition of equality of matrices we get,

⇒ 2x - y = 10          (...Eq 1)

⇒ 3x = 6 or x = 2.

Putting value of x in Eq 1

⇒ 2x - y = 10
⇒ 2(2) - y = 10
⇒ 4 - y = 10
⇒ y = 4 - 10
⇒ y = -6

∴ x = 2 and y = -6.

∴ Option 2 is the correct option.

$\text{If A }= \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*},$ then A² =

1. $\begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*}$

2. $\begin{bmatrix*}[r] 0 & 0 \\ 1 & 1 \end{bmatrix*}$

3. $\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}$

4. $\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$

Answer

Given,

$\text{A} = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.$

∴ Option 4 is the correct option.

$\text{If A }= \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*},$ then A² =

1. A
2. O
3. I
4. 2A

Answer

Given,

$\text{A} = \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 0 \times 1 & 0 \times 0 + 0 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 0 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \therefore \text{A}^2 = O.$

∴ Option 2 is the correct option.

$\text{If A }= \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*},$ then A² =

1. $\begin{bmatrix*}[r] 2 & 0 \\ 1 & 1 \end{bmatrix*}$

2. $\begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}$

3. $\begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}$

4. none of these

Answer

Given,

$\text{A} = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} \\[0.5em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times 0 + 1 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} \\[0.5em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}.$

∴ Option 3 is the correct option.

$\text{If A }= \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*},$ then A² =

1. $\begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*}$

2. $\begin{bmatrix*}[r] 8 & -5 \\ 5 & 3 \end{bmatrix*}$

3. $\begin{bmatrix*}[r] 8 & -5 \\ -5 & -3 \end{bmatrix*}$

4. $\begin{bmatrix*}[r] 8 & -5 \\ -5 & 3 \end{bmatrix*}$

Answer

Given,

$\text{A} = \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ (-1) \times 3 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*} \\[0.5em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*}.$

∴ Option 1 is the correct option.

If matrix 𝐴 = $\begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*}\text{ and } A^2 = \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*}$, then the value of x is :

1. 2

2. 4

3. 8

4. 10

Answer

$\phantom{\Rightarrow} A^2 = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow A^2 = \begin{bmatrix*}[r] 2 \times 2 + 2\times 0 & 2 \times 2 + 2 \times 2 \\ 0 \times 2 + 2 \times 0 & 0 \times 2 + 2 \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 4 + 4 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 8 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow x = 8.$

Hence, Option 3 is the correct option.

If A = $\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*},$ then A² = pA, then the value of p is

1. 2

2. 4

3. -2

4. -4

Answer

Given,

$\text{A}^2 = pA \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + (-2) \times (-2) & 2 \times (-2) + (-2) \times 2 \\ -2 \times 2 + 2 \times (-2) & (-2) \times (-2) + 2 \times 2 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 4 + 4 & -4 - 4 \\ -4 - 4 & 4 + 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 8 & -8 \\ -8 & 8 \end{bmatrix*} = \begin{bmatrix*}[r] 2p & -2p \\ -2p & 2p \end{bmatrix*} \\[0.5em]$

By definition of equality of matrices we get,

⇒ 2p = 8

∴ p = 4.

∴ Option 2 is the correct option.

The product AB of two matrices A and B is possible if

1. A and B have same number of rows.

2. A and B have same number of columns.

3. The number of columns of A is equal to the number of rows of B.

4. The number of rows of A is equal to the number of columns of B.

Answer

Matrix multiplication is possible if the number of columns of the first matrix is the same as the number of rows as the second matrix.

So, AB is possible if the number of columns of A is equal to the number of rows of B.

Hence, option 3 is the correct option.

If A = $\begin{bmatrix*}[r] -1 & 2 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*}$. Which of the following operations is possible ?

1. A - B

2. A + B

3. AB

4. BA

Answer

Order of A : 1 × 2 and order of B : 2 × 2

Since, no. of columns in A is same as the no. of rows in B.

∴ AB is possible.

Hence, Option 3 is the correct option.