If $A=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right], B=\left[\begin{array}{cc}4 & 0 \\ 1 & 5\end{array}\right], C=\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right], a=4$, and $b=-2$, then show that
(i) $A+(B+C)=(A+B)+C$
(ii) $A(B C)=(A B) C$
(iii) $(a+b) B=a B+b B$
(iv) $a(C-A)=a C-a A$
(v) $\left(A^T\right)^T=A$
(vi) $(b A)^T=b A^T$
(vii) $(A B)^T=B^T A^T$
(viii) $(A-B) C=A C-B C$
(ix) $(A-B)^T=A^T-B^T$
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,
$$\begin{aligned} & A C=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 4 & -4 \\ 1 & -6 \end{array}\right] \quad\text{.... (ii)}\\ \text{and}\quad & B C=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right]\quad\text{.... (iii)} \end{aligned}$$
$$\begin{aligned} \therefore \quad A C-B C & =\left[\begin{array}{cc} 4-8 & -4-0 \\ 1-7 & -6+10 \end{array}\right] \quad \text{[using Eqs. (ii) and (iii)]}\\ & =\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right] \\ & =(A-B) C\quad\text{[using Eq. (i)] Hence proved.} \end{aligned}$$
$$\begin{aligned} \text { (ix) } \quad&(A-B)^T=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]^{\top} \\ &=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]^{\top}=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right] \\ & A^T-B^T=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]-\left[\begin{array}{cc} 4 & 1 \\ 0 & 5 \end{array}\right] \\ &=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right]=(A-B)^T\quad\text{Hence proved.} \end{aligned}$$
If $A=\left[\begin{array}{cc}\cos q & \sin q \\ -\sin q & \cos q\end{array}\right]$, then show that $A^2=\left[\begin{array}{cc}\cos 2 q & \sin 2 q \\ -\sin 2 q & \cos 2 q\end{array}\right]$.
$$\begin{aligned} & \text { We have, } A=\left[\begin{array}{cc} \cos q & \sin q \\ -\sin q & \cos q \end{array}\right] \\ & \therefore \quad A^2=A \cdot A=\left[\begin{array}{cc} \cos q & \sin q \\ -\sin q & \cos q \end{array}\right] \cdot\left[\begin{array}{cc} \cos q & \sin q \\ -\sin q & \cos q \end{array}\right] \end{aligned}$$
$$\begin{aligned} & =\left[\begin{array}{cc} \cos ^2 q-\sin ^2 q & \cos q \cdot \sin q+\sin q \cos q \\ -\sin q \cos q-\cos q \sin q & -\sin ^2 q+\cos ^2 q \end{array}\right] \\ & =\left[\begin{array}{cc} \cos 2 q & 2 \sin q \cos q \\ -2 \sin q \cos q & \cos 2 q \end{array}\right] \quad\left[\because \cos ^2 \theta-\sin ^2 \theta=\cos 2 \theta\right] \\ & =\left[\begin{array}{cc} \cos 2 q & \sin 2 q \\ -\sin 2 q & \cos 2 q \end{array}\right] \quad[\because \sin 2 \theta=2 \sin \theta \cdot \cos \theta] \text { Hence proved. } \end{aligned}$$
If $A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ and $x^2=-1$, then show that $(A+B)^2=A^2+B^2$.
$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} A & =\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right], B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \text { and } x^2=-1 \\ \therefore\quad (A+B) & =\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right] \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { and } \quad (A+B)^2 & =\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right] \\ & =\left[\begin{array}{cc} 1-x^2 & 0 \\ 0 & 1-x^2 \end{array}\right]\quad\text{.... (i)} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { Also, }\quad & A^2=A \cdot A=\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]=\left[\begin{array}{cc} -x^2 & 0 \\ 0 & -x^2 \end{array}\right] \\ \text{and}\quad & B^2=B \cdot B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned} \end{aligned}$$
$$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} A^2+B^2 & =\left[\begin{array}{cc} -x^2+1 & 0 \\ 0 & -x^2+1 \end{array}\right]=\left[\begin{array}{cc} 1-x^2 & 0 \\ 0 & 1-x^2 \end{array}\right] \quad\text{using Eq. (i)]}\\ & =(A+B)^2\quad\text{Hence proved.} \end{aligned} \end{aligned}$$
Verify that $A^2=I$, when $A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right]$.
We have, $A=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right]$
$\therefore\quad A^2=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right] \cdot\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right] \quad\left[\because A^2=A \cdot A\right]$
$$=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I$$ Hence proved.
Prove by mathematical induction that $\left(A^{\prime}\right)^n=\left(A^n\right)^{\prime}$ where $n \in N$ for any square matrix $A$.
Let $$P(n):\left(A^n\right)^n=\left(A^n\right)^{\prime}$$
$\therefore \quad P(1):(A)^1=(A)^{\prime}$
$\Rightarrow \quad A^{\prime}=A^{\prime} \Rightarrow P(1)$ is true.
Now, $$\quad P(k):(A)^k=\left(A^k\right)^{\prime}$$
where $k \in N$
and $$\quad P(k+1):\left(A^{\prime}\right)^{k+1}=\left(A^{k+1}\right)^{\prime}$$
$$\begin{aligned} &\text { where } P(k+1) \text { is true whenever } P(k) \text { is true. }\\ &\begin{aligned} \therefore \quad P(k+1):\left(A^{\prime}\right)^k \cdot\left(A^{\prime}\right)^{\prime} & =\left[A^{k+1}\right]^{\prime} \\ \left(A^k\right)^{\prime} \cdot(A)^{\prime} & =\left[A^{k+1}\right]^{\prime} \\ \left(A \cdot A^k\right)^{\prime} & =\left[A^{k+1}\right]^{\prime} \quad \left[\because(A)^{\prime k}=\left(A^k\right)^{\prime} \text { and }(A B)=B^{\prime} A^{\prime}\right]\\ \left(A^{k+1}\right)^{\prime} & =\left[A^{k+1}\right]^{\prime}\text{Hence proved.} \end{aligned} \end{aligned}$$