Exercise-3.3, Class 12th, Maths, Chapter 3, NCERT

Q1. (i)

Matrix $A=\left[\begin{array}{c}5\\ \frac{1}{2}\\ -1\end{array}\right]$

This is a column matrix (3×1).

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Transpose:

To find $A^{\mathrm{\prime }}$(or $A^T$), we interchange rows and columns.

So,

$$A^{\mathrm{\prime }}=[5\frac{1}{2}-1]$$

Answer:

$$A^{\mathrm{\prime }}=[5\frac{1}{2}-1]$$

(This is now a row matrix (1×3).)

 

Q1- (ii).

Matrix $B=\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]$

This is a 2×2 matrix.

Transpose:

We swap rows with columns:

$$B^{\mathrm{\prime }}=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$$

Answer:

$$B^{\mathrm{\prime }}=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$$

 

Q1 (iii)

Matrix

$$A=\left[\begin{array}{ccc}-1& 5& 6\\ \sqrt{3}& 5& 6\\ 2& 3& -1\end{array}\right]$$

This is a 3×3 matrix (3 rows, 3 columns).

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To find: $A^{\mathrm{\prime }}$ (the transpose)

To find the transpose, we interchange rows and columns — that is, the first row becomes the first column, the second row becomes the second column, and so on.

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Step-by-step:

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Transpose:

$$A^{\mathrm{\prime }}=\left[\begin{array}{ccc}-1& \sqrt{3}& 2\\ 5& 5& 3\\ 6& 6& -1\end{array}\right]$$

 

$$A^{\mathrm{\prime }}=[5\frac{1}{2}-1]$$

(This is now a row matrix (1×3).)

 

Question 2.

Given:

$$A=\left[\begin{array}{ccc}-1& 2& 3\\ 5& 7& 9\\ -2& 1& 1\end{array}\right],B=\left[\begin{array}{ccc}-4& 1& -5\\ 1& 2& 0\\ 1& 3& 1\end{array}\right]$$

We have to verify:

1️⃣ $(A+B{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}+B^{\mathrm{\prime }}$

2️⃣ $(A-B{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}-B^{\mathrm{\prime }}$

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Step 1: Find A + B

Add corresponding elements:

$$A+B=\left[\begin{array}{ccc}(-1)+(-4)& 2+1& 3+(-5)\\ 5+1& 7+2& 9+0\\ (-2)+1& 1+3& 1+1\end{array}\right]=\left[\begin{array}{ccc}-5& 3& -2\\ 6& 9& 9\\ -1& 4& 2\end{array}\right]$$

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Now find (A + B)′

Take the transpose — interchange rows and columns:

$$(A+B{)}^{\mathrm{\prime }}=\left[\begin{array}{ccc}-5& 6& -1\\ 3& 9& 4\\ -2& 9& 2\end{array}\right]$$

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Step 2: Find A′ and B′ separately

Transpose of A:

$$A^{\mathrm{\prime }}=\left[\begin{array}{ccc}-1& 5& -2\\ 2& 7& 1\\ 3& 9& 1\end{array}\right]$$

Transpose of B:

$$B^{\mathrm{\prime }}=\left[\begin{array}{ccc}-4& 1& 1\\ 1& 2& 3\\ -5& 0& 1\end{array}\right]$$

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Step 3: Find A′ + B′

$$A^{\mathrm{\prime }}+B^{\mathrm{\prime }}=\left[\begin{array}{ccc}(-1)+(-4)& 5+1& (-2)+1\\ 2+1& 7+2& 1+3\\ 3+(-5)& 9+0& 1+1\end{array}\right]=\left[\begin{array}{ccc}-5& 6& -1\\ 3& 9& 4\\ -2& 9& 2\end{array}\right]$$

This is exactly equal to $(A+B{)}^{\mathrm{\prime }}$.

Hence,

$$\boxed{{\displaystyle (A+B{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}+B^{\mathrm{\prime }}}}$$

Step 4: Now find A – B

Subtract corresponding elements:

$$A-B=\left[\begin{array}{ccc}(-1)-(-4)& 2-1& 3-(-5)\\ 5-1& 7-2& 9-0\\ (-2)-1& 1-3& 1-1\end{array}\right]=\left[\begin{array}{ccc}3& 1& 8\\ 4& 5& 9\\ -3& -2& 0\end{array}\right]$$

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Now find (A – B)′

$$(A-B{)}^{\mathrm{\prime }}=\left[\begin{array}{ccc}3& 4& -3\\ 1& 5& -2\\ 8& 9& 0\end{array}\right]$$

Step 5: Find A′ – B′

$$A^{\mathrm{\prime }}-B^{\mathrm{\prime }}=\left[\begin{array}{ccc}(-1)-(-4)& 5-1& (-2)-1\\ 2-1& 7-2& 1-3\\ 3-(-5)& 9-0& 1-1\end{array}\right]=\left[\begin{array}{ccc}3& 4& -3\\ 1& 5& -2\\ 8& 9& 0\end{array}\right]$$

This is exactly equal to $(A-B{)}^{\mathrm{\prime }}$.

Hence,

$$\boxed{{\displaystyle (A-B{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}-B^{\mathrm{\prime }}}}$$

 

Question 3.

Given

$A^{\mathrm{\prime }}=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]$

, then verify that

(i) (A + B)′ = A′ + B′ 

(ii) (A – B)′ = A′ – B′

 Determine $A$ from $A^{\mathrm{\prime }}$

Remember: $A^{\mathrm{\prime }}$ is the transpose of $A$.
So to get $A$, we transpose $A^{\mathrm{\prime }}$ again.

$$A=(A^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right]$$

Hence

$$A=\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right],B=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]$$

Compute $A+B$

Add corresponding elements:

$$A+B=\left[\begin{array}{ccc}3+(-1)& -1+2& 0+1\\ 4+1& 2+2& 1+3\end{array}\right]=\left[\begin{array}{ccc}2& 1& 1\\ 5& 4& 4\end{array}\right]$$

 Compute $(A+B{)}^{\mathrm{\prime }}$

Transpose the above matrix:

$$(A+B{)}^{\mathrm{\prime }}=\left[\begin{array}{cc}2& 5\\ 1& 4\\ 1& 4\end{array}\right]$$

 Compute $A^{\mathrm{\prime }}+B^{\mathrm{\prime }}$

We already know:

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right],B^{\mathrm{\prime }}=\left[\begin{array}{cc}-1& 1\\ 2& 2\\ 1& 3\end{array}\right]$$

Now add elementwise:

$$A^{\mathrm{\prime }}+B^{\mathrm{\prime }}=\left[\begin{array}{cc}3+(-1)& 4+1\\ -1+2& 2+2\\ 0+1& 1+3\end{array}\right]=\left[\begin{array}{cc}2& 5\\ 1& 4\\ 1& 4\end{array}\right]$$

Therefore

$$\boxed{{\displaystyle (A+B{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}+B^{\mathrm{\prime }}}}$$

 Compute $A-B$

$$A-B=\left[\begin{array}{ccc}3-(-1)& -1-2& 0-1\\ 4-1& 2-2& 1-3\end{array}\right]=\left[\begin{array}{ccc}4& -3& -1\\ 3& 0& -2\end{array}\right]$$ Compute $(A-B{)}^{\mathrm{\prime }}$

$$(A-B{)}^{\mathrm{\prime }}=\left[\begin{array}{cc}4& 3\\ -3& 0\\ -1& -2\end{array}\right]$$

 Compute $A^{\mathrm{\prime }}-B^{\mathrm{\prime }}$

$$A^{\mathrm{\prime }}-B^{\mathrm{\prime }}=\left[\begin{array}{cc}3-(-1)& 4-1\\ -1-2& 2-2\\ 0-1& 1-3\end{array}\right]=\left[\begin{array}{cc}4& 3\\ -3& 0\\ -1& -2\end{array}\right]$$

Therefore

$$\boxed{{\displaystyle (A-B{)}^{\mathrm{\prime }}=A^{\mathrm{\prime }}-B^{\mathrm{\prime }}}}$$

Question 4.

Given:

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}-2& 3\\ 1& 2\end{array}\right],B=\left[\begin{array}{cc}-1& 0\\ 1& 2\end{array}\right]$$

We need to find

$$(A+2B{)}^{\mathrm{\prime }}$$

 Find $A$ from $A^{\mathrm{\prime }}$

Since $A^{\mathrm{\prime }}=\left[\begin{array}{cc}-2& 3\\ 1& 2\end{array}\right]$,

Transpose it to get $A$:

$$A=(A^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\left[\begin{array}{cc}-2& 1\\ 3& 2\end{array}\right]$$

 Compute $2B$

Multiply each element of $B$ by 2:

$$2B=\left[\begin{array}{cc}2\times (-1)& 2\times 0\\ 2\times 1& 2\times 2\end{array}\right]=\left[\begin{array}{cc}-2& 0\\ 2& 4\end{array}\right]$$

 Compute $A+2B$

$$A+2B=\left[\begin{array}{cc}-2& 1\\ 3& 2\end{array}\right]+\left[\begin{array}{cc}-2& 0\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}-4& 1\\ 5& 6\end{array}\right]$$

 Find $(A+2B{)}^{\mathrm{\prime }}$

Take the transpose:

$$(A+2B{)}^{\mathrm{\prime }}=\left[\begin{array}{cc}-4& 5\\ 1& 6\end{array}\right]$$

Question 5.

For the matrices A and B, verify that (AB)′ = B′A′, where

Concept Recap

For any conformable matrices $A$ and $B$:

$$(AB{)}^{\mathrm{\prime }}=B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}$$

That is, the transpose of a product = product of transposes in reverse order.

Now, let’s check numerically.

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(i)

$$A=\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right],B=\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]$$

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Step 1️⃣: Compute $AB$

Since $A$ is 3×1 and $B$ is 1×3,
$AB$ will be a 3×3 matrix.

$$AB=\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]=\left[\begin{array}{ccc}1(-1)& 1(2)& 1(1)\\ -4(-1)& -4(2)& -4(1)\\ 3(-1)& 3(2)& 3(1)\end{array}\right]=\left[\begin{array}{ccc}-1& 2& 1\\ 4& -8& -4\\ -3& 6& 3\end{array}\right]$$

 Find $(AB{)}^{\mathrm{\prime }}$

Transpose means interchange rows and columns:

$$(AB{)}^{\mathrm{\prime }}=\left[\begin{array}{ccc}-1& 4& -3\\ 2& -8& 6\\ 1& -4& 3\end{array}\right]$$

 Compute $B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}$

Now find $B^{\mathrm{\prime }}$ and $A^{\mathrm{\prime }}$.

$$B^{\mathrm{\prime }}=\left[\begin{array}{c}-1\\ 2\\ 1\end{array}\right],A^{\mathrm{\prime }}=\left[\begin{array}{ccc}1& -4& 3\end{array}\right]$$

Multiply $B^{\mathrm{\prime }}$ (3×1) with $A^{\mathrm{\prime }}$ (1×3):

$$B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}=\left[\begin{array}{c}-1\\ 2\\ 1\end{array}\right]\left[\begin{array}{ccc}1& -4& 3\end{array}\right]=\left[\begin{array}{ccc}(-1)(1)& (-1)(-4)& (-1)(3)\\ (2)(1)& (2)(-4)& (2)(3)\\ (1)(1)& (1)(-4)& (1)(3)\end{array}\right]$$

$$=\left[\begin{array}{ccc}-1& 4& -3\\ 2& -8& 6\\ 1& -4& 3\end{array}\right]$$

Hence,

$$(AB{)}^{\mathrm{\prime }}=B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}$$

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(ii)

$$A=\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right],B=\left[\begin{array}{ccc}1& 5& 7\end{array}\right]$$

Compute $AB$

$$AB=\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]\left[\begin{array}{ccc}1& 5& 7\end{array}\right]=\left[\begin{array}{ccc}0(1)& 0(5)& 0(7)\\ 1(1)& 1(5)& 1(7)\\ 2(1)& 2(5)& 2(7)\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 1& 5& 7\\ 2& 10& 14\end{array}\right]$$

 Find $(AB{)}^{\mathrm{\prime }}$

$$(AB{)}^{\mathrm{\prime }}=\left[\begin{array}{ccc}0& 1& 2\\ 0& 5& 10\\ 0& 7& 14\end{array}\right]$$

Compute $B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}$

$$B^{\mathrm{\prime }}=\left[\begin{array}{c}1\\ 5\\ 7\end{array}\right],A^{\mathrm{\prime }}=\left[\begin{array}{ccc}0& 1& 2\end{array}\right]$$

Multiply:

$$B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}=\left[\begin{array}{c}1\\ 5\\ 7\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\end{array}\right]=\left[\begin{array}{ccc}1(0)& 1(1)& 1(2)\\ 5(0)& 5(1)& 5(2)\\ 7(0)& 7(1)& 7(2)\end{array}\right]=\left[\begin{array}{ccc}0& 1& 2\\ 0& 5& 10\\ 0& 7& 14\end{array}\right]$$

Hence,

$$(AB{)}^{\mathrm{\prime }}=B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}$$$$\boxed{{\displaystyle {\mathrm{}}^{}}}$$

 

Question 6.

verify that A′ A = I

(i)

$$A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$$

 Find $A^{\mathrm{\prime }}$

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$$

Compute $A^{\mathrm{\prime }}A$

$$A^{\mathrm{\prime }}A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$$
$$A^{\mathrm{\prime }}A=\left[\begin{array}{cc}({\cos }^2\alpha +{\sin }^2\alpha )& (\cos \alpha \sin \alpha -\sin \alpha \cos \alpha )\\ (\sin \alpha \cos \alpha -\cos \alpha \sin \alpha )& ({\sin }^2\alpha +{\cos }^2\alpha )\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$$

Hence, $A^{\mathrm{\prime }}A=I$

(ii)

$$A=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right]$$

 Find $A^{\mathrm{\prime }}$

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]$$

 Compute $A^{\mathrm{\prime }}A$

$$A^{\mathrm{\prime }}A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right]$$
$$A^{\mathrm{\prime }}A=\left[\begin{array}{cc}({\sin }^2\alpha +{\cos }^2\alpha )& (\sin \alpha \cos \alpha -\cos \alpha \sin \alpha )\\ (\cos \alpha \sin \alpha -\sin \alpha \cos \alpha )& ({\cos }^2\alpha +{\sin }^2\alpha )\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$$

Hence, $A^{\mathrm{\prime }}A=I$

Q7.

Recall:

• A matrix $A$ is symmetric if $A^{\mathrm{\prime }}=A$.

• A matrix $A$ is skew-symmetric if $A^{\mathrm{\prime }}=-A$. (Note diagonal entries of a skew-symmetric matrix must be $0$.)

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(i) Show $A$ is symmetric

$$A=\left[\begin{array}{ccc}1& -1& 5\\ -1& 2& 1\\ 5& 1& 3\end{array}\right]$$

Find the transpose $A^{\mathrm{\prime }}$ (swap rows and columns):

$$A^{\mathrm{\prime }}=\left[\begin{array}{ccc}1& -1& 5\\ -1& 2& 1\\ 5& 1& 3\end{array}\right]$$

We see $A^{\mathrm{\prime }}=A$.
Therefore $A$ is symmetric. 

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(ii) Show $A$ is skew-symmetric

$$A=\left[\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ 1& -1& 0\end{array}\right]$$

Compute the transpose:

$$A^{\mathrm{\prime }}=\left[\begin{array}{ccc}0& -1& 1\\ 1& 0& -1\\ -1& 1& 0\end{array}\right]$$

Compute $-A$:

$$-A=\left[\begin{array}{ccc}0& -1& 1\\ 1& 0& -1\\ -1& 1& 0\end{array}\right]$$

We have $A^{\mathrm{\prime }}=-A$. Also all diagonal entries are $0$, as required.

Therefore $A$ is skew-symmetric. 

 

Question 8.

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

Given

$$A=\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]\mathrm{.}$$

First find the transpose:

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]\mathrm{.}$$

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(i) $A+A^{\mathrm{\prime }}$

Compute:

$$A+A^{\mathrm{\prime }}=\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]+\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]=\left[\begin{array}{cc}2& 11\\ 11& 14\end{array}\right]\mathrm{}$$

Take its transpose:

$$(A+A^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\left[\begin{array}{cc}2& 11\\ 11& 14\end{array}\right]\mathrm{}$$Since $(A+A^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=A+A^{\mathrm{\prime }}$, the matrix $A+A^{\mathrm{\prime }}$ is symmetric.

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(ii) $A-A^{\mathrm{\prime }}$

Compute:

$$A-A^{\mathrm{\prime }}=\left[\begin{array}{cc}1& 5\\ 6& 7\end{array}\right]-\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\mathrm{}$$

Take its transpose:

$$(A-A^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=-(A-A^{\mathrm{\prime }})\mathrm{}$$

Since $(A-A^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=-(A-A^{\mathrm{\prime }})$, the matrix $A-A^{\mathrm{\prime }}$ is skew-symmetric.

 

Question 9. Find

$$\frac{1}{2}(A+A^{\mathrm{\prime }})\text{and}\frac{1}{2}(A-A^{\mathrm{\prime }})$$

We are given:

$$A=\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]$$

 Find $A^{\mathrm{\prime }}$

Transpose means interchange rows and columns:

$$A^{\mathrm{\prime }}=\left[\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right]$$ Compute $A+A^{\mathrm{\prime }}$

Add corresponding entries of $A$ and $A^{\mathrm{\prime }}$:

$$A+A^{\mathrm{\prime }}=\left[\begin{array}{ccc}0+0& a+(-a)& b+(-b)\\ (-a)+a& 0+0& c+(-c)\\ (-b)+b& (-c)+c& 0+0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

 Compute $A-A^{\mathrm{\prime }}$

Subtract corresponding entries:

$$A-A^{\mathrm{\prime }}=\left[\begin{array}{ccc}0-0& a-(-a)& b-(-b)\\ (-a)-a& 0-0& c-(-c)\\ (-b)-b& (-c)-c& 0-0\end{array}\right]=\left[\begin{array}{ccc}0& 2a& 2b\\ -2a& 0& 2c\\ -2b& -2c& 0\end{array}\right]$$

Multiply by ½

$$\frac{1}{2}(A+A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$
$$\frac{1}{2}(A-A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{ccc}0& 2a& 2b\\ -2a& 0& 2c\\ -2b& -2c& 0\end{array}\right]=\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]$$

Final Answers

$$\boxed{{\displaystyle \frac{1}{2}(A+A^{\mathrm{\prime }})=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],\frac{1}{2}(A-A^{\mathrm{\prime }})=\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]}}$$

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

we’ll write each matrix $A$ as

$$A=\frac{1}{2}(A+A^{\mathrm{\prime }})+\frac{1}{2}(A-A^{\mathrm{\prime }}),$$

where $S=\frac{1}{2}(A+A^{\mathrm{\prime }})$ is symmetric and $K=\frac{1}{2}(A-A^{\mathrm{\prime }})$ is skew-symmetric. I’ll give $S$ and $K$ for each part.

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(i) $A=\left[\begin{array}{cc}3& 5\\ 1& -1\end{array}\right]$

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}3& 1\\ 5& -1\end{array}\right]$$
$$S=\frac{1}{2}(A+A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{cc}6& 6\\ 6& -2\end{array}\right]=\left[\begin{array}{cc}3& 3\\ 3& -1\end{array}\right]$$

$$K=\frac{1}{2}(A-A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{cc}0& 4\\ -4& 0\end{array}\right]=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$$

So $A=S+K$ with $S$ symmetric and $K$ skew.

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(ii) $A=\left[\begin{array}{ccc}6& -2& 2\\ -2& 3& -1\\ 2& -1& 3\end{array}\right]$

This matrix is already symmetric (check $A^{\mathrm{\prime }}=A$), so

$$S=\frac{1}{2}(A+A^{\mathrm{\prime }})=A=\left[\begin{array}{ccc}6& -2& 2\\ -2& 3& -1\\ 2& -1& 3\end{array}\right],K=\frac{1}{2}(A-A^{\mathrm{\prime }})=\mathbf{0}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\mathrm{.}$$

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(iii) $A=\left[\begin{array}{ccc}3& 3& -1\\ -2& -2& 1\\ -4& -5& 2\end{array}\right]$

First compute $A^{\mathrm{\prime }}=\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

Then

$$S=\frac{1}{2}(A+A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{ccc}6& 1& -5\\ 1& -4& -4\\ -5& -4& 4\end{array}\right]=\left[\begin{array}{ccc}3& \frac{1}{2}& -\frac{5}{2}\\ \frac{1}{2}& -2& -2\\ -\frac{5}{2}& -2& 2\end{array}\right],$$

which is symmetric, and

$$K=\frac{1}{2}(A-A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{ccc}0& 5& 3\\ -5& 0& 6\\ -3& -6& 0\end{array}\right]=\left[\begin{array}{ccc}0& \frac{5}{2}& \frac{3}{2}\\ -\frac{5}{2}& 0& 3\\ -\frac{3}{2}& -3& 0\end{array}\right],<span\; class="minner"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="mord"><span\; class="mtable"><span\; class="col-align-c"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span></span></span></span></span><span\; class="mclose"><span\; class="delimsizing\; mult"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span></span></span></span></span>$$

which is skew-symmetric. (You can check $S^{\mathrm{\prime }}=S$ and $K^{\mathrm{\prime }}=-K$)

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(iv) $A=\left[\begin{array}{cc}1& 5\\ -1& 2\end{array}\right]$

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}1& -1\\ 5& 2\end{array}\right]$$
$$S=\frac{1}{2}(A+A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{cc}2& 4\\ 4& 4\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 2& 2\end{array}\right],$$

$$K=\frac{1}{2}(A-A^{\mathrm{\prime }})=\frac{1}{2}\left[\begin{array}{cc}0& 6\\ -6& 0\end{array}\right]=\left[\begin{array}{cc}0& 3\\ -3& 0\end{array}\right]\mathrm{.}$$

Question 11

Given:
$A$ and $B$ are symmetric matrices of the same order.
That means:

$$A^{\mathrm{\prime }}=A\text{and}{B}^{\mathrm{\prime }}=B$$

We need to find the nature of the matrix $AB-BA$

 Take transpose of $AB-BA$

$$(AB-BA{)}^{\mathrm{\prime }}=(AB{)}^{\mathrm{\prime }}-(BA{)}^{\mathrm{\prime }}=B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}-A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}$$

Since $A^{\mathrm{\prime }}=A$ and $B^{\mathrm{\prime }}=B$,

$$(AB-BA{)}^{\mathrm{\prime }}=BA-AB=-(AB-BA)$$

 Interpretation

A matrix $M$ is skew-symmetric if $M^{\mathrm{\prime }}=-M$.

Here we found $(AB-BA{)}^{\mathrm{\prime }}=-(AB-BA)$ Therefore, $AB-BA$ is a skew-symmetric matrix.

Correct Option: (A) Skew symmetric matrix

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Question 12

Given:

$$A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right],\text{and}A+A^{\mathrm{\prime }}=I$$

We must find the value of $\alpha$

Compute $A^{\mathrm{\prime }}$

$$A^{\mathrm{\prime }}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$$

 Compute $A+A^{\mathrm{\prime }}$

$$A+A^{\mathrm{\prime }}=\left[\begin{array}{cc}\cos \alpha +\cos \alpha & -\sin \alpha +\sin \alpha \\ \sin \alpha -\sin \alpha & \cos \alpha +\cos \alpha \end{array}\right]=\left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]$$

 Given $A+A^{\mathrm{\prime }}=I$

$$\left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

This gives:

$$2\cos \alpha =1⟹\cos \alpha =\frac{1}{2}$$

 Solve for $\alpha$

$$\cos \alpha =\frac{1}{2}\Rightarrow \alpha =\frac{\pi }{3}$$

 Final Answers 
1️⃣ $AB-BA$ → Skew-symmetric matrix
2️⃣ $A+A^{\mathrm{\prime }}=I\Rightarrow \alpha ={\displaystyle \frac{\pi }{3}}$