Matrices — Multiple Choice Questions

The order of the matrix having x rows and y columns is denoted by:

1. x + y

2. x - y

3. x × y

4. none of these

Answer

The order of a matrix is written as "rows by columns."

If a matrix has x rows and y columns, its order is written as x × y.

Hence, option 3 is the correct option.

The total number of elements in an m × n matrix is:

1. m + n

2. mⁿ

3. mn

4. 2^{m + n}

Answer

The order of an m × n matrix indicates that it has m rows and n columns. To find the total number of elements, you multiply the number of rows by the number of columns = mn.

Hence, option 3 is the correct option.

The matrix, A = $\begin{bmatrix} 3 & 4 & -8 \\ 2 & 0 & -2 \\ 1 & 7 & -9 \end{bmatrix}$, then the value of a₂₃ is:

1. 2

2. -8

3. -9

4. -2

Answer

i = Row

j = Column

In a₂₃, i = 2 (Second row) j = 3 (Third column)

In given matrix:

A = $\begin{bmatrix} 3 & 4 & -8 \\ 2 & 0 & -2 \\ 1 & 7 & -9 \end{bmatrix}$

The element in the second row and third column is −2.

Hence, option 4 is the correct option.

The matrix, A = $\begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -7 \end{bmatrix}$ is a:

1. Rectangular matrix

2. Column matrix

3. Diagonal matrix

4. Identity matrix

Answer

Given,

A = $\begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -7 \end{bmatrix}$

A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero.

Thus, A is a diagonal matrix.

Hence, option 3 is the correct option.

A matrix of order of 3 × 1 whose elements a_ij are given by a_ij = (2i + j), is :

1. $\begin{bmatrix} 3 & 5 & 7 \end{bmatrix}$

2. $\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$

3. $\begin{bmatrix} 2 & 4 & 6 \end{bmatrix}$

4. $\begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

Answer

a_ij = (2i + j)

⇒ a₁₁ = 2(1) + 1 = 3

⇒ a₂₁ = (2)2 + 1 = 5

⇒ a₃₁ = (2)3 + 1 = 7

Matrix A = $\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$

Hence, option 2 is the correct option.

A matrix of order 3 × 2 whose elements a_ij are given by a_ij = (i + j), is:

1. $\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix}$

2. $\begin{bmatrix} 2 & 3 & 4 \\ 4 & 5 & 6 \end{bmatrix}$

3. $\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 4 & 5 \end{bmatrix}$

4. $\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$

Answer

a_ij = i + j.

∴ a₁₁ = 1 + 1 = 2, a₁₂ = 1 + 2 = 3.

a₂₁ = 2 + 1 = 3, a₂₂ = 2 + 2 = 4.

a₃₁ = 3 + 1 = 4, a₃₂ = 3 + 2 = 5.

Matrix A = $\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 4 & 5 \end{bmatrix}$

Hence, option 3 is the correct option.

If a matrix has 12 elements, then the total number of possible orders it can have is:

1. 4

2. 6

3. 8

4. cannot be determined

Answer

If a matrix has 12 elements the possible orders are, 1 x 12, 12 x 1, 2 x 6, 6 x 2, 3 x 4, 4 x 3.

There are 6 possible orders.

Hence, option 2 is the correct option.

A 2 × 2 matrix whose elements are given by a_ij = $\dfrac{(i + 2j)^2}{2}$ is:

1. $\begin{bmatrix} \dfrac{3}{2} & \dfrac{5}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}$

2. $\begin{bmatrix} \dfrac{5}{2} & \dfrac{7}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}$

3. $\begin{bmatrix} \dfrac{9}{2} & \dfrac{15}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}$

4. $\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}$

Answer

Given,

a_ij = $\dfrac{(i + 2j)^2}{2}$

$a_{11} = \dfrac{[1 + 2(1)]^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}, a_{12} = \dfrac{[1 + 2(2)]^2}{2} = \dfrac{5^2}{2} = \dfrac{25}{2} \\[1em] a_{21} = \dfrac{[2 + 2]^2}{2} = \dfrac{4^2}{2} = \dfrac{16}{2} = 8, a_{22} = \dfrac{[2 + 2(2)]^2}{2} = \dfrac{6^2}{2} = \dfrac{36}{2} = 18.$

Matrix A = $\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\ 8 & 18 \end{bmatrix}$

Hence, option 4 is the correct option.

If $\begin{bmatrix} 5 & 3 \\ x & 7 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 7 \end{bmatrix}$, then the value of (x + y + z) is:

1. 9

2. 8

3. 11

4. 10

Answer

Given,

$\begin{bmatrix} 5 & 3 \\ x & 7 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 7 \end{bmatrix}$

Solving for x, y and z:

∴ x = 1

∴ y = 5

∴ z = 3

⇒ x + y + z = 9

Hence, option 1 is the correct option.

If $\begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$, then the value of (x + y + z + w) is:

1. 8

2. 9

3. 10

4. 12

Answer

Given,

$\Rightarrow \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$

Solving for x and y:

∴ 2x - y = 0

⇒ y = 2x ...(1)

∴ x - y = -1 ...(2)

Substituting value of y from equation (1) in x - y = -1, we get:

⇒ x - 2x = -1

⇒ -x = -1

⇒ x = 1.

Substituting value of x in equation(1), we get:

⇒ y = 2(1)

⇒ y = 2.

Solving for w and z:

∴ 2x + z = 5

⇒ 2(1) + z = 5

⇒ 2 + z = 5

⇒ z = 5 - 2

⇒ z = 3.

∴ 3z + w = 13

⇒ 3(3) + w = 13

⇒ 9 + w = 13

⇒ w = 13 - 9

⇒ w = 4.

∴ x + y + z + w = 1 + 2 + 3 + 4 = 10.

Hence, option 3 is the correct option.

Which of the following statements is true for the transpose of a matrix?

1. The number of rows is same as that of the given matrix.

2. The number of columns is same as that of the given matrix.

3. The order is same as that of the given matrix.

4. The number of elements is same as that of the given matrix.

Answer

If A = m × n then A^T = n × m

The total number of elements is calculated by multiplying the dimensions: m × n. Since m × n = n × m

Therefore, total number of elements in A is exactly the same as the total number of elements in A^T

Hence, option 4 is the correct option.

If $\begin{bmatrix} x + y \\ x - y \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix}$, then the value of xy is:

1. 4

2. 8

3. 10

4. 12

Answer

Given,

$\begin{bmatrix} x + y \\ x - y \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix}$

∴ x + y = 8 .....(1)

∴ x - y = 4 .....(2)

Adding equations (1) and (2), we get :

⇒ x + y + x - y = 8 + 4

⇒ 2x = 12

⇒ x = $\dfrac{12}{2}$

⇒ x = 6.

Substituting value of x in equation (1) :

⇒ x + y = 8

⇒ 6 + y = 8

⇒ y = 8 - 6

⇒ y = 2.

∴ xy = 6 × 2 = 12.

Hence, option 4 is the correct option.

If $\begin{bmatrix} p - q & 2q \\ 2q + r & p + q \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix}$, then the value of (p + q + r) is:

1. 8

2. 10

3. -5

4. -10

Answer

$\begin{bmatrix} p - q & 2q \\ 2q + r & p + q \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix}$

Solving for p, q and r:

∴ 2q = 4...(1)

⇒ q = $\dfrac{4}{2}$

⇒ q = 2.

∴ p - q = 1

⇒ p - 2 = 1

⇒ p = 1 + 2

⇒ p = 3.

∴ 2q + r = 9

⇒ 2(2) + r = 9

⇒ 4 + r = 9

⇒ r = 9 - 4

⇒ r = 5.

∴ p + q + r = 3 + 2 + 5 = 10.

Hence, option 2 is the correct option.

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

1. 2, 7

2. 7, 2

3. 3, 5

4. 2, -7

Answer

Given,

$\begin{bmatrix} x + 3 & 4 \\ y - 4 & x + y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}$

Solving for x and y :

∴ x + 3 = 5

⇒ x = 5 - 3

⇒ x = 2.

∴ y - 4 = 3

⇒ y = 3 + 4

⇒ y = 7.

Hence, option 1 is the correct option.

$\begin{bmatrix} 2 & 4 & 1 & 4 & 2 \\ 1 & -2 & 3 & 5 & 1 \\ 3 & 1 & 2 & 3 & 6 \end{bmatrix}$ is a matrix of the order:

1. 3 × 4

2. 5 × 3

3. 3 × 5

4. none of these

Answer

Given matrix has :

3 rows and 5 columns

Therefore, the order of the matrix is 3 × 5.

Hence, option 3 is the correct option.

If matrix A of order 2 × 1 and matrix B of order 2 × 2 are added, then the order of the matrix A + B is:

1. 2 × 2

2. 2 × 1

3. 1 × 2

4. A + B is not possible

Answer

Given,

A of order 2 × 1

B of order 2 × 2

Since their orders are different, thus matrix A and B cannot be added.

Thus, A + B is not possible.

Hence, option 4 is the correct option.

If A = $\begin{bmatrix} -5 & 4 \\ 3 & -8 \end{bmatrix}$ and B = $\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}$, then (A – B) is:

1. $\begin{bmatrix} 7 & 7 \\ -4 & 12 \end{bmatrix}$

2. $\begin{bmatrix} -7 & 7 \\ 4 & -12 \end{bmatrix}$

3. $\begin{bmatrix} -7 & -7 \\ 4 & 12 \end{bmatrix}$

4. $\begin{bmatrix} 3 & 1 \\ 2 & -4 \end{bmatrix}$

Answer

Given,

A = $\begin{bmatrix} -5 & 4 \\ 3 & -8 \end{bmatrix}$ and B = $\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}$

Solving for A - B:

$\Rightarrow \begin{bmatrix} -5 - 2 & 4 - (-3) \\ 3 - (-1) & -8 - 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 & 7 \\ 4 & -12 \end{bmatrix}.$

Hence, option 1 is the correct option.

If A = $\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix}$, then the value of (-3)A is:

1. $\begin{bmatrix} -8 & -4 \\ 6 & 18 \end{bmatrix}$

2. $\begin{bmatrix} -12 & -6 \\ 9 & 18 \end{bmatrix}$

3. $\begin{bmatrix} -12 & -6 \\ 9 & 27 \end{bmatrix}$

4. $\begin{bmatrix} -12 & 6 \\ 9 & -27 \end{bmatrix}$

Answer

Given,

$\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix}$

Solving for -3A:

$\Rightarrow -3\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -12 & -6 \\ 9 & 27 \end{bmatrix}$

Hence, option 3 is the correct option.

If A = $\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix}$ and B = $\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix}$, then (3A + 2B) is equal to:

1. $\begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}$

2. $\begin{bmatrix} 8 & 2 \\ 7 & 3 \end{bmatrix}$

3. $\begin{bmatrix} 19 & 16 \\ 18 & 17 \end{bmatrix}$

4. $\begin{bmatrix} 9 & 16 \\ 8 & 17 \end{bmatrix}$

Answer

Given,

A = $\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix}$ and B = $\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix}$

Solving for 3A + 2B:

$\Rightarrow 3\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 & 6 \\ 12 & 3 \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 6 & 4 \end{bmatrix} \\[1em] \Rightarrow 3\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 + 10 & 6 + 0 \\ 12 + 6 & 3 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}$

Hence, option 1 is the correct option.

Find the matrix A, if A + $\begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix}$ = $\begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix}$

1. $\begin{bmatrix} -1 & 12 \\ 8 & -15 \end{bmatrix}$

2. $\begin{bmatrix} -1 & -12 \\ 8 & -15 \end{bmatrix}$

3. $\begin{bmatrix} -1 & -12 \\ -8 & -15 \end{bmatrix}$

4. $\begin{bmatrix} 1 & 12 \\ 8 & 15 \end{bmatrix}$

Answer

Given,

A + $\begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix}$ = $\begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix}$

Solving for A:

$\Rightarrow A = \begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - 4 & -6 - 6 \\ 5 - (-3) & -8 - 7 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 & -12 \\ 8 & -15 \end{bmatrix}$

Hence, option 2 is the correct option.

$\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \\ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \\ \cosθ & \sinθ \end{bmatrix}$ is equal to:

1. $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

2. $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

3. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

4. none of these

Answer

Given,

$\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \\ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \\ \cosθ & \sinθ \end{bmatrix}$

Solving:

$\Rightarrow \begin{bmatrix} \cos^2θ & \cosθ\sinθ \\ -\sinθ\cosθ & \cos^2θ \end{bmatrix} + \cdot \begin{bmatrix} \sin^2θ & -\sinθ\cosθ \\ \sinθ\cosθ & \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2θ + \sin^2θ & \cosθ\sinθ - \sinθ\cosθ\ -\sinθ\cosθ + \sinθ\cosθ & \cos^2θ + \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$

Hence, option 3 is the correct option.

If $2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$, then the value of (x – y) is:

1. 10

2. -10

3. 6

4. -6

Answer

Given,

$2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$

Solving:

$\Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}.$

∴ 8 + y = 0

⇒ y = -8

∴ 2x + 1 = 5

⇒ 2x = 5 - 1

⇒ 2x = 4

⇒ x = $\dfrac{4}{2}$

⇒ x = 2

∴ x - y = 2 - (-8) = 10

Hence, option 1 is the correct option.

If $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$,then the value of (x + y) is:

1. 4

2. -4

3. 6

4. 8

Answer

Given,

$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$

Solving:

$\Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.$

∴ 2 + y = 5

⇒ y = 5 - 2

⇒ y = 3

∴ 2x + 2 = 8

⇒ 2x = 8 - 2

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$ = 3.

∴ x + y = 3 + 3 = 6.

Hence, option 3 is the correct option.

If $\begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$, then the values of a and b are:

1. a = 4, b = 2

2. a = 2, b = 4

3. both (a) and (b)

4. none of these

Answer

Given,

$\begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$

∴ a + b = 6

⇒ b = 6 - a...(1)

∴ ab = 8...(2)

Substituting value of b from equation(1) in ab = 8, we get:

⇒ a (6 - a) = 8

⇒ 6a - a² = 8

⇒ a² - 6a + 8 = 0

⇒ a² - 4a - 2a + 8 = 0

⇒ a(a - 4) -2(a - 4) = 0

⇒ (a - 2)(a - 4) = 0

(a - 2)= 0 or (a - 4) = 0 [Using zero product rule]

⇒ a = 2 or a = 4

If a = 2, then b = 6 − 2 = 4.

If a = 4, then b = 6 − 4 = 2.

Hence, option 3 is the correct option.

If A = $\begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$, then (A + A′) is equal to :

1. 0

2. 2A

3. –A′

4. A′

Answer

Given,

A = $\begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$

A′ = $\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}$

(A + A′)

$\Rightarrow \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 + 0 & -1 + 1 & 2 + (-2) \\ 1 + (-1) & 0 + 0 & 3 + (-3) \\ -2 + 2 & -3 + 3 & 0 + 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Hence, option 1 is the correct option.

If $\begin{bmatrix} x & y + 2 & z - 3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$, then the value of xyz is:

1. 24

2. 20

3. 36

4. 30

Answer

$\Rightarrow \begin{bmatrix} x & y + 2 & z - 3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + y & y + 2 + 4 & z - 3 + 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + y & y + 6 & z + 2 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em]$

Solving for y and z:

∴ y + 6 = 9

⇒ y = 9 - 6

⇒ y = 3.

∴ x + y = 4

⇒ x + 3 = 4

⇒ x = 4 - 3

⇒ x = 1

∴ z + 2 = 12

⇒ z = 12 - 2

⇒ z = 10.

∴ xyz = (1)(3)(10) = 30.

Hence, option 4 is the correct option.

If $x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix}$ = 0, then the value of (2x – 3y) is:

1. 1

2. -4

3. 3

4. -3

Answer

$\Rightarrow x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ x \end{bmatrix} + \begin{bmatrix} 3y \\ 5y \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 3y - 8 \\ x + 5y - 11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$

∴ 2x + 3y - 8 = 0

⇒ 2x + 3y = 8 ....(1)

∴ x + 5y - 11 = 0

⇒ x = 11 - 5y ....(2)

Substituting value of x from equation (2) in (1), we get :

⇒ 2(11 - 5y) + 3y = 8

⇒ 22 - 10y + 3y = 8

⇒ -7y = 8 - 22

⇒ -7y = -14

⇒ y = $\dfrac{-14}{-7}$

⇒ y = 2.

Substituting value of y in equation (2) we get:

⇒ x = 11 - 5(2)

⇒ x = 11 - 10

⇒ x = 1.

∴ 2x - 3y = 2(1) - 3(2) = 2 - 6 = -4.

Hence, option 2 is the correct option.

If $\begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}$, then the matrix A is:

1. $\begin{bmatrix} -3 & 5 \\ 8 & 15 \end{bmatrix}$

2. $\begin{bmatrix} 5 & -3 \\ 15 & 8 \end{bmatrix}$

3. $\begin{bmatrix} -3 & \dfrac{5}{2} \\ 8 & \dfrac{15}{2} \end{bmatrix}$

4. $\begin{bmatrix} \dfrac{5}{2} & -3 \\ \dfrac{15}{2} & 8 \end{bmatrix}$

Answer

Given,

$\Rightarrow \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(4\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(\begin{bmatrix} -4 & 8 \\ 12 & 16 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(\begin{bmatrix} -4 - 2 & 8 - 3 \\ 12 - (-4) & 16 - 1 \end{bmatrix}\Big) \\[1em] \Rightarrow \dfrac{1}{2}\begin{bmatrix} -6 & 5 \\ 16 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 & \dfrac{5}{2} \\ 8 & \dfrac{15}{2} \end{bmatrix}.$

Hence, option 3 is the correct option.

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

1. A and B have the same number of rows.

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

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

4. A and B have the same number of columns.

Answer

For matrix multiplication A × B to be defined, the inner dimensions must match.

If matrix A has the order m × n.And matrix B has the order p × q.The product AB is only possible if n = p.

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

Hence, option 2 is the correct option.

If the orders of matrices A and B are m × n and p × q respectively, then which of the following is true for the product BA?

1. m = p

2. m = q

3. n = p

4. n = q

Answer

Given,

Matrix A: m × n

Matrix B: p × q

The product BA is defined if number of columns in B = number of rows in A, m = q .

Hence, option 2 is the correct option.

If P and Q are two different matrices of order 3 × n and n × p, then the order of the matrix PQ is:

1. 3 × p

2. p × 3

3. n × n

4. 3 × 3

Answer

Given,

Matrix P: 3 × n

Matrix Q: n × p

The product PQ is defined if number of columns in P × number of rows in Q, 3 × p .

Hence, option 1 is the correct option.

If $\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \times P = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$, then the order of matrix P is:

1. 2 × 2

2. 2 × 1

3. 1 × 2

4. cannot be determined

Answer

Let A = $\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$

Let B = $\begin{bmatrix} 6 \\ 8 \end{bmatrix}$

Let order of P be m × n:

Thus, AP = B

For matrix multiplication:

The number of columns in A must equal the number of rows in P.

⇒ m = 2

The order of the product B = Rows of A × Columns of P.

⇒ 2 × 1 = 2 × n

⇒ n = 1.

Order of matrix P = 2 × 1.

Hence, option 2 is the correct option.

If A = $\begin{bmatrix} 3 & -6 \end{bmatrix}$ and B = $\begin{bmatrix} 3 & -1 \\ 4 & 3 \end{bmatrix}$, then the order of matrix AB is:

1. 1 × 2

2. 2 × 1

3. 2 × 2

4. 1 × 3

Answer

A = $\begin{bmatrix} 3 & -6 \end{bmatrix}$

Order of matrix A = 1 × 2

B = $\begin{bmatrix} 3 & -1 \\ 4 & 3 \end{bmatrix}$

Order of matrix B = 2 × 2

The resulting matrix AB will have the number of rows from the first matrix A and the number of columns from the second matrix B.

Order of matrix AB = 1 × 2

Hence, option 1 is the correct option.

If $\begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ = $\begin{bmatrix} 2 \\ -8 \end{bmatrix}$, the values of x and y respectively are:

1. 1, -2

2. -2, 1

3. 1, 2

4. -2, -1

Answer

Given,

$\begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ = $\begin{bmatrix} 2 \\ -8 \end{bmatrix}$

Solving,

$\Rightarrow \begin{bmatrix} 2(x) + 0(y) \\ 0(x) + 4(y) \end{bmatrix}= \begin{bmatrix} 2 \\ -8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ 4y \end{bmatrix}= \begin{bmatrix} 2 \\ -8 \end{bmatrix}$

∴ 2x = 2

⇒ x = $\dfrac{2}{2}$

⇒ x = 1.

∴ 4y = -8

⇒ y = $\dfrac{-8}{4}$

⇒ y = -2.

Hence, option 1 is the correct option.

If $\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = 0$, then the value of x is :

1. 0

2. 1

3. 2

4. -2

Answer

Given,

$\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = 0$

Solving,

$\Rightarrow \begin{bmatrix} x(1) + (1)(-2) & (x)(0) + (1)(0) \end{bmatrix} = 0 \\[1em] \Rightarrow \begin{bmatrix} x - 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$

∴ x - 2 = 0

⇒ x = 2

Hence, option 3 is the correct option.

If matrix A = $\begin{bmatrix} -1 & 2 \end{bmatrix}$ and matrix B = $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$, then matrix AB is equal to:

1. $\begin{bmatrix} -3 \end{bmatrix}$

2. $\begin{bmatrix} 8 \end{bmatrix}$

3. $\begin{bmatrix} 5 \end{bmatrix}$

4. $\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}$

Answer

Calculating,

$\Rightarrow AB = \begin{bmatrix} -1 & 2 \\ \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 \times 3 + 2 \times 4 \\ \end{bmatrix} \\[1em] = \begin{bmatrix} -3 + 8 \\ \end{bmatrix} \\[1em] = \begin{bmatrix} 5 \end{bmatrix}.$

Hence, option 3 is the correct option.

If, $x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$, then the value of xy is:

1. 3

2. -12

3. -6

4. 15

Answer

Solving for x and y:

$\Rightarrow x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em]$

∴ 2x - y = 10 ....(1)

∴ 3x + y = 5 .....(2)

Adding equations (1) and (2), we get :

⇒ 2x - y + 3x + y = 10 + 5

⇒ 2x + 3x = 15

⇒ 5x = 15

⇒ x = $\dfrac{15}{5}$

⇒ x = 3.

Substituting value of x in 2x - y = 10, we get :

⇒ 2(3) - y = 10

⇒ 6 - y = 10

⇒ y = 6 - 10

⇒ y = -4.

∴ xy = (3)(-4) = -12.

Hence, option 2 is the correct option.

If, A = $\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}$ and B = $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$, then the matrix AB is :

1. $\begin{bmatrix} -7 \\ 2 \end{bmatrix}$

2. $\begin{bmatrix} -7 & 2 \end{bmatrix}$

3. $\begin{bmatrix} -7 \\ -2 \end{bmatrix}$

4. $\begin{bmatrix} -7 & -2 \end{bmatrix}$

Answer

Given,

A = $\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}$ and B = $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$

Solving for AB:

$\Rightarrow \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} -4 \\ 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (4)(-4) + (3)(3) \\ (1)(-4) + (2)(3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -16 + 9 \\ -4 + 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 \\ 2 \end{bmatrix}.$

Hence, option 1 is the correct option.

If A = $\begin{bmatrix} a & b \end{bmatrix}$ and B = $\begin{bmatrix} c \\ d \end{bmatrix}$, then :

1. only matrix AB is possible

2. only matrix BA is possible

3. both matrices AB and BA are possible

4. both matrices AB and BA are possible, AB = BA

Answer

Order of matrix A = 1 × 2

Order of matrix B = 2 × 1

Since, no. of columns in A is equal to the no. of rows in B and no. of columns in B is equal to the no. of rows in A.

∴ AB and BA are possible.

$AB = \begin{bmatrix} a \times c + b \times d \end{bmatrix} \\[1em] = \begin{bmatrix} ac + bd \end{bmatrix}. \\[1em] BA = \begin{bmatrix} c \times a & c \times b \\ d \times a & d \times b \end{bmatrix} \\[1em] = \begin{bmatrix} ca & cb \\ da & db \end{bmatrix}.$

∴ AB ≠ BA.

Hence, Option 3 is the correct option.

If matrix A = $\begin{bmatrix*}[r] -1 & 2 \end{bmatrix*}$ and matrix B = $\begin{bmatrix*}[r] 3 \\ 4 \end{bmatrix*}$, then matrix AB is equal to:

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

$2. \begin{bmatrix*}[r] 8 \end{bmatrix*}$

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

$4. \begin{bmatrix*}[r] -1 & 2 \\ 3 & 4 \end{bmatrix*}$

Answer

Calculating,

$\Rightarrow AB = \begin{bmatrix*}[r] -1 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 3 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 3 + 2 \times 4 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 + 8 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \end{bmatrix*}.$

Hence, option 3 is the correct option.

If A is a matrix of order m × n such that AB and BA are both defined, then B is a/an:

1. m × n matrix

2. n × m matrix

3. m × m matrix

4. n × n matrix

Answer

If A is of order m × n:

For matrix multiplication AB to exist, the number of columns of A must equal the number of rows of B.

Number of rows in B = n

For matrix multiplication BA to exist, the number of columns of B must be equal to the number of rows of A.

Number of columns in B = m.

B is of order = n × m.

Hence, option 2 is the correct option.

If A = $\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}$ and I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then AI is equal to:

1. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

2. $\begin{bmatrix} 5 & 10 \\ -3 & 4 \end{bmatrix}$

3. $\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}$

4. $\begin{bmatrix} 15 & 15 \\ -1 & -1 \end{bmatrix}$

Answer

Given,

A = $\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}$ and I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Solving for AI:

$AI = \begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 5(1) + 10(0) & 5(0) + 10(1) \\ 3(1) + (-4)(0) & 3(0) + (-4)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 5 + 0 & 0 + 10 \\ 3 + 0 & 0 + (-4) \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}.$

Hence, option 3 is the correct option.

If $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, then A² is equal to:

1. $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$

2. $\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$

3. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

4. $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

Answer

Given,

$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

Substituting values we get :

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