$$\begin{aligned} \text{Let}\quad P & =A^{\prime} A \\ \therefore\quad P^{\prime} & =\left(A A^{\prime}\right)^{\prime} \\ & =A^{\prime}\left(A^{\prime}\right)^{\prime} \quad \left[\because\left(A B^{\prime}\right)^{\prime}=B^{\prime} A^{\prime}\right]\\ & =A^{\prime} A=P \end{aligned}$$

So, $A^{\prime} A$ is symmetric matrix for any matrix $A$.

Similarly, let $$Q=A A^{\prime}$$

$$\begin{array}{l} \therefore \quad Q^{\prime} & =\left(A A^{\prime}\right)^{\prime}=\left(A^{\prime}\right)^{\prime}(A)^{\prime} \\ & =A\left(A^{\prime}\right)^{\prime}=Q \end{array}$$

So, AA' is symmetric matrix for any matrix A.

$$\begin{aligned} &\text { Since, } A \text { and } B \text { are square matrices of order } 3 \times 3 \text {. }\\ &\therefore \quad A B^2=A B \cdot A B \end{aligned}$$

$$\begin{aligned} & =A B A B \\ & =A A B B \\ & =A^2 B^2 \end{aligned} \quad[\because A B=B A]$$

So, $A B^2=A^2 B^2$ is true when $A B=B A$.

$$\begin{aligned} &\text { Since, } A \text { and } B \text { are square matrices such that } A B=B A \text {. }\\ &\begin{aligned} \therefore \quad(A+B)^2 & =(A+B) \cdot(A+B) \\ & =A^2+A B+B A+B^2 \\ & =A^2+A B+A B+B^2 \quad [\because A B=B A]\\ & =A^2+2 A B+B^2\quad \text{Hence proved.} \end{aligned} \end{aligned}$$

We have,

$$\begin{aligned} & A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right] \\ & C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right] \text { and } a=4, b=-2 \end{aligned}$$

$$\begin{aligned} &\text { (i) }\\ &\begin{aligned} & A+(B+C)=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left[\begin{array}{ll} 6 & 0 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right] \\ & \text { and } \\ & (A+B)+C=\left[\begin{array}{ll} 5 & 2 \\ 0 & 8 \end{array}\right]+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right] \\ & =\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]=A+(B+C)\quad\text{Hence proved.} \end{aligned} \end{aligned}$$

$$\begin{aligned} & \text { (ii) }(B C)=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right] \\ & \text { and } \\ & A(B C)=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right] \\ & =\left[\begin{array}{cc} 8+14 & 0-20 \\ -8+21 & 0-30 \end{array}\right]=\left[\begin{array}{ll} 22 & -20 \\ 13 & -30 \end{array}\right] \end{aligned}$$

$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} (A B) & =\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right] \cdot\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{cc} 6 & 10 \\ -1 & 15 \end{array}\right] \\ (A B) C & =\left[\begin{array}{cc} 6 & 10 \\ -1 & 15 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right] \\ & =\left[\begin{array}{cc} 22 & -20 \\ 13 & -30 \end{array}\right]=A(B C)\quad\text{Hence proved.} \end{aligned} \end{aligned}$$

(iii) $(a+b) B=(4-2)\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\quad$ $$[\because a=4, b=-2]$$

$$\begin{aligned} & =\left[\begin{array}{cc} 8 & 0 \\ 2 & 10 \end{array}\right] \\ \text{and}\quad a B+b B & =4 B-2 B \\ & =\left[\begin{array}{cc} 16 & 0 \\ 4 & 20 \end{array}\right]-\left[\begin{array}{cc} 8 & 0 \\ 2 & 10 \end{array}\right] \\ & =\left[\begin{array}{cc} 8 & 0 \\ 2 & 10 \end{array}\right] \\ & =(a+b) B\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\text { (iv) }\\ &\begin{aligned} & (C-A)=\left[\begin{array}{cc} 2-1 & 0-2 \\ 1+1 & -2-3 \end{array}\right]=\left[\begin{array}{ll} 1 & -2 \\ 2 & -5 \end{array}\right] \\ & \text { and } \\ & a(C-A)=\left[\begin{array}{ll} 4 & -8 \\ 8 & -20 \end{array}\right]\quad [\because a=4] \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { Also, } \quad a C-a A & =\left[\begin{array}{cc} 8 & 0 \\ 4 & -8 \end{array}\right]-\left[\begin{array}{cc} 4 & 8 \\ -4 & 12 \end{array}\right]=\left[\begin{array}{cc} 4 & -8 \\ 8 & -20 \end{array}\right] \\ & =a(C-A)\quad \text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\text { (v) }\\ &\begin{aligned} A^T=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^T & =\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]^{\top} \\ \text { Now, } \quad\left(A^T\right)^T & =\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^T \\ & =A\quad\text{Hence proved.} \end{aligned} \end{aligned}$$

(vi) $(b A)^{\top}=\left[\begin{array}{cc}-2 & -4 \\ 2 & -6\end{array}\right]^{\top}\quad$ $$[\because b=-2]$$

$$\begin{aligned} & =\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right] \\ \text{and}\quad A^T & =\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right] \\ \therefore\quad b A^T & =\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]=(b A)^T\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} & \text { (vii) } A B=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{cc} 4+2 & 0+10 \\ -4+3 & 0+15 \end{array}\right]=\left[\begin{array}{cc} 6 & 10 \\ -1 & 15 \end{array}\right] \\ & \therefore \quad(A B)^T=\left[\begin{array}{cc} 6 & -1 \\ 10 & 15 \end{array}\right] \\ & \text { Now, } \\ & B^T A^T=\left[\begin{array}{ll} 4 & 1 \\ 0 & 5 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} 6 & -1 \\ 10 & 15 \end{array}\right] \\ & =(A B)^T\quad\text{Hence proved.} \end{aligned}$$

$$\text { (viii) } \begin{aligned} &(A-B)=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right] \\ &(A-B) C=\left[\begin{array}{lc} -3 & 2 \\ -2 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right]\quad\text{.... (i)} \end{aligned}$$

Now,

$$\begin{aligned} & A C=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 4 & -4 \\ 1 & -6 \end{array}\right] \quad\text{.... (ii)}\\ \text{and}\quad & B C=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right]\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} \therefore \quad A C-B C & =\left[\begin{array}{cc} 4-8 & -4-0 \\ 1-7 & -6+10 \end{array}\right] \quad \text{[using Eqs. (ii) and (iii)]}\\ & =\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right] \\ & =(A-B) C\quad\text{[using Eq. (i)] Hence proved.} \end{aligned}$$

$$\begin{aligned} \text { (ix) } \quad&(A-B)^T=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]^{\top} \\ &=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]^{\top}=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right] \\ & A^T-B^T=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]-\left[\begin{array}{cc} 4 & 1 \\ 0 & 5 \end{array}\right] \\ &=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right]=(A-B)^T\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} & \text { We have, } A=\left[\begin{array}{cc} \cos q & \sin q \\ -\sin q & \cos q \end{array}\right] \\ & \therefore \quad A^2=A \cdot A=\left[\begin{array}{cc} \cos q & \sin q \\ -\sin q & \cos q \end{array}\right] \cdot\left[\begin{array}{cc} \cos q & \sin q \\ -\sin q & \cos q \end{array}\right] \end{aligned}$$

$$\begin{aligned} & =\left[\begin{array}{cc} \cos ^2 q-\sin ^2 q & \cos q \cdot \sin q+\sin q \cos q \\ -\sin q \cos q-\cos q \sin q & -\sin ^2 q+\cos ^2 q \end{array}\right] \\ & =\left[\begin{array}{cc} \cos 2 q & 2 \sin q \cos q \\ -2 \sin q \cos q & \cos 2 q \end{array}\right] \quad\left[\because \cos ^2 \theta-\sin ^2 \theta=\cos 2 \theta\right] \\ & =\left[\begin{array}{cc} \cos 2 q & \sin 2 q \\ -\sin 2 q & \cos 2 q \end{array}\right] \quad[\because \sin 2 \theta=2 \sin \theta \cdot \cos \theta] \text { Hence proved. } \end{aligned}$$