Rbse Solutions Class 12 Maths: Exercise 3.3

This exercise focuses on the properties of the Transpose of a Matrix ($A’$ or $A^T$), and the definition and application of Symmetric and Skew Symmetric Matrices.

Question 1. Find the transpose of each of the following matrices:

(i) $A = \begin{bmatrix} 5 \\ 1/2 \\ -1 \end{bmatrix}$ (ii) $B = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ (iii) $C = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & 1 \end{bmatrix}$

Solution:

The transpose is found by interchanging rows and columns ($A’ = [a_{ji}]$).

(i) $A’ = \begin{bmatrix} 5 & 1/2 & -1 \end{bmatrix}$

(ii) $B’ = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

(iii) $C’ = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & 1 \end{bmatrix}$

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Question 2.

If $A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$, verify that:

(i) $(A + B)’ = A’ + B’$ (ii) $(A – B)’ = A’ – B’$

Solution:

First, find the transposes $A’$ and $B’$:

$$A’ = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}, \quad B’ = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}$$

(i) Verify $(A + B)’ = A’ + B’$ (Transpose of sum is sum of transposes)

• LHS: $(A + B)’$$$A+B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} \implies (A+B)’ = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$$
• RHS: $A’ + B’$$$A’ + B’ = \begin{bmatrix} -1-4 & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3-5 & 9+0 & 1+1 \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$$Since LHS = RHS, the property is verified.

(ii) Verify $(A – B)’ = A’ – B’$

• LHS: $(A – B)’$$$A-B = \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5) \\ 5-1 & 7-2 & 9-0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$$$$(A-B)’ = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$$
• RHS: $A’ – B’$$$A’ – B’ = \begin{bmatrix} -1-(-4) & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3-(-5) & 9-0 & 1-1 \end{bmatrix} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$$Since LHS = RHS, the property is verified.

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Question 3.

If $A’ = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$, verify that:

(i) $(A + B)’ = A’ + B’$ (ii) $(A – B)’ = A’ – B’$

Solution:

First, find $A$ and $B’$:

$$A = (A’)’ = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}, \quad B’ = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$$

(i) Verify $(A + B)’ = A’ + B’$

• LHS: $(A + B)’$$$A+B = \begin{bmatrix} 3-1 & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$$$$(A+B)’ = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$$
• RHS: $A’ + B’$$$A’ + B’ = \begin{bmatrix} 3-1 & 4+1 \\ -1+2 & 2+2 \\ 0+1 & 1+3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$$Since LHS = RHS, the property is verified.

(ii) Verify $(A – B)’ = A’ – B’$

• LHS: $(A – B)’$$$A-B = \begin{bmatrix} 3-(-1) & -1-2 & 0-1 \\ 4-1 & 2-2 & 1-3 \end{bmatrix} = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}$$$$(A-B)’ = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$$
• RHS: $A’ – B’$$$A’ – B’ = \begin{bmatrix} 3-(-1) & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$$Since LHS = RHS, the property is verified.

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Question 4.

If $A’ = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, then find $(A + 2B)’$.

Solution:

1. Find $A$ from $A’$:$$A = (A’)’ = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$$
2. Calculate $2B$:$$2B = 2 \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$$
3. Calculate $A + 2B$:$$A + 2B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -2-2 & 1+0 \\ 3+2 & 2+4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$$
4. Find the transpose $(A + 2B)’$:$$(A + 2B)’ = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$$

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Question 5.

For the matrices $A$ and $B$, verify that $(AB)’ = B’A’$, where:

(i) $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$

Solution (i):

• LHS: $(AB)’$$$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1(-1) & 1(2) & 1(1) \\ -4(-1) & -4(2) & -4(1) \\ 3(-1) & 3(2) & 3(1) \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$$$$(AB)’ = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$$
• RHS: $B’A’$$$A’ = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}, \quad B’ = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$$$$B’A’ = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1(1) & -1(-4) & -1(3) \\ 2(1) & 2(-4) & 2(3) \\ 1(1) & 1(-4) & 1(3) \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$$Since LHS = RHS, the property $(AB)’ = B’A’$ is verified.

(ii) $A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$

Solution (ii):

• LHS: $(AB)’$$$AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$$$$(AB)’ = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$$
• RHS: $B’A’$$$A’ = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}, \quad B’ = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$$$$B’A’ = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 1(0) & 1(1) & 1(2) \\ 5(0) & 5(1) & 5(2) \\ 7(0) & 7(1) & 7(2) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$$Since LHS = RHS, the property $(AB)’ = B’A’$ is verified.

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Question 6.

If:

(i) $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, then verify that $A’A = I$.

(ii) $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$, then verify that $A’A = I$.

Solution:

(i) Verify $A’A = I$

$$A’ = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$$

$$A’A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha – \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha – \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix}$$

Using $\cos^2 \alpha + \sin^2 \alpha = 1$:

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

The property is verified.

(ii) Verify $A’A = I$

$$A’ = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$$

$$A’A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} = \begin{bmatrix} \sin^2 \alpha + \cos^2 \alpha & \sin \alpha \cos \alpha – \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha – \sin \alpha \cos \alpha & \cos^2 \alpha + \sin^2 \alpha \end{bmatrix}$$

Using $\sin^2 \alpha + \cos^2 \alpha = 1$:

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

The property is verified. (This type of matrix is known as an Orthogonal Matrix).

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Question 7.

(i) Show that the matrix $A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.

(ii) Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$ is a skew symmetric matrix.

Solution:

(i) Symmetric Matrix ($A’ = A$):

$$A’ = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$$

Since $A’ = A$, the matrix $A$ is symmetric.

(ii) Skew Symmetric Matrix ($A’ = -A$):

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

Now calculate $-A$:

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

Since $A’ = -A$, the matrix $A$ is skew symmetric. (Note that the main diagonal elements are all zero, which is a condition for skew symmetric matrices).

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Question 8.

For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that:

(i) $(A + A’)$ is a symmetric matrix.

(ii) $(A – A’)$ is a skew symmetric matrix.

Solution:

First find $A’$:

$$A’ = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$$

(i) $P = A + A’$ (Symmetric check: $P’ = P$)

$$P = A + A’ = \begin{bmatrix} 1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$$

$$P’ = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$$

Since $P’ = P$, $(A + A’)$ is a symmetric matrix.

(ii) $Q = A – A’$ (Skew Symmetric check: $Q’ = -Q$)

$$Q = A – A’ = \begin{bmatrix} 1-1 & 5-6 \\ 6-5 & 7-7 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

$$Q’ = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

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

Since $Q’ = -Q$, $(A – A’)$ is a skew symmetric matrix.

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Question 9.

Find $\frac{1}{2}(A + A’)$ and $\frac{1}{2}(A – A’)$, when $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.

Solution:

First find $A’$:

$$A’ = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$$

1. Calculate $A + A’$:$$A+A’ = \begin{bmatrix} 0+0 & a-a & b-b \\ -a+a & 0+0 & c-c \\ -b+b & -c+c & 0+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$$$\frac{1}{2}(A + A’) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$The sum of a skew symmetric matrix and its transpose is always the zero matrix.
2. Calculate $A – A’$:$$A-A’ = \begin{bmatrix} 0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{bmatrix} = \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}$$$$\frac{1}{2}(A – A’) = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$$This result is simply the matrix A itself, which is expected since A is already skew symmetric.

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Question 10.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ (iv) $A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$

Note: Questions (ii) and (iii) are similar and omitted for brevity, but the method is identical.

Solution:

A matrix $A$ can be uniquely expressed as $A = P + Q$, where $P = \frac{1}{2}(A + A’)$ (Symmetric) and $Q = \frac{1}{2}(A – A’)$ (Skew Symmetric).

(i) $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$

$$A’ = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$$

1. Symmetric part $P = \frac{1}{2}(A + A’)$:$$A+A’ = \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} \implies P = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$$
2. Skew Symmetric part $Q = \frac{1}{2}(A – A’)$:$$A-A’ = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} \implies Q = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$$$\text{Sum: } \mathbf{\begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}}$$

(iv) $A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$

$$A’ = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$$

1. Symmetric part $P = \frac{1}{2}(A + A’)$:$$A+A’ = \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} \implies P = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}$$
2. Skew Symmetric part $Q = \frac{1}{2}(A – A’)$:$$A-A’ = \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} \implies Q = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$$$$\text{Sum: } \mathbf{\begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}}$$

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Question 11.

If $A, B$ are symmetric matrices of same order, then $AB – BA$ is a:

(A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix

Solution:

Given: $A’ = A$ and $B’ = B$.

We need to check the nature of $P = AB – BA$.

$$P’ = (AB – BA)’$$

Using $(M – N)’ = M’ – N’$ and $(MN)’ = N’M’$:

$$P’ = (AB)’ – (BA)’ = B’A’ – A’B’$$

Substitute $A’ = A$ and $B’ = B$:

$$P’ = BA – AB$$

$$P’ = -(AB – BA) = -P$$

Since $P’ = -P$, $AB – BA$ is a skew symmetric matrix.

$$\text{The correct option is } \mathbf{(A)}$$

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Question 12.

If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, and $A + A’ = I$, then the value of $\alpha$ is:

(A) $\pi/6$ (B) $\pi/3$ (C) $\pi$ (D) $3\pi/2$

Solution:

1. Find $A + A’$:$$A’ = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$$$$A + A’ = \begin{bmatrix} \cos \alpha + \cos \alpha & -\sin \alpha + \sin \alpha \\ \sin \alpha – \sin \alpha & \cos \alpha + \cos \alpha \end{bmatrix} = \begin{bmatrix} 2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha \end{bmatrix}$$
2. Equate to $I$:Given $A + A’ = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.$$\begin{bmatrix} 2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
3. Solve for $\alpha$:$2 \cos \alpha = 1 \implies \cos \alpha = 1/2$The principal value for which $\cos \alpha = 1/2$ is $\alpha = \pi/3$.

$$\text{The correct option is } \mathbf{(B)}$$

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