If $A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$, then verify that
(i) $\left(A^{\prime}\right)^{\prime}=A$
(ii) $(A B)^{\prime}=B^{\prime} A^{\prime}$
(iii) $(k A)^{\prime}=\left(k A^{\prime}\right)$.
We have, $A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$
(i) We have to verify that, $A^{\prime}=A$
$$\begin{aligned} \therefore\quad & A^{\prime}=\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right] \\ \text{and}\quad & A^{\prime}=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]=A\quad\text{Hence proved.} \end{aligned}$$
$$\begin{aligned} &\text { (ii) We have to verify that, } A B^{\prime}=B^{\prime} A^{\prime}\\ &\begin{array}{ll} \therefore & A B=\left[\begin{array}{cc} 3 & 9 \\ 11 & -15 \end{array}\right] \\ \Rightarrow & (A B)^{\prime}=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \end{array} \end{aligned}$$
$$\begin{array}{rlr} \text{and}\quad B^{\prime} A^{\prime} & =\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{rr} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \\ & =(A B)^{\prime} \quad \text { Hence proved. } \end{array}$$
(iii) We have to verify that, $(k A)^{\prime}=\left(k A^{\prime}\right)$
$$\begin{aligned} \text{Now,}\quad (k A) & =\left[\begin{array}{ccc} 0 & -k & 2 k \\ 4 k & 3 k & -4 k \end{array}\right] \\ \text{and}\quad (k A)^{\prime} & =\left[\begin{array}{cc} 0 & 4 k \\ -k & 3 k \\ 2 k & -4 k \end{array}\right] \\ \text{Also,}\quad k A^{\prime} & =\left[\begin{array}{cc} 0 & 4 k \\ -k & 3 k \\ 2 k & -4 k \end{array}\right] \\ & =(k A)^{\prime}\quad\text{Hence proved.} \end{aligned}$$
If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$, then verify that
(i) $(2 A+B)^{\prime}=2 A \mathrm{~A}+B^{\prime}$.(ii) $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.
We have, $$ A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]$$
(i) $\therefore(2 A+B)=\left[\begin{array}{cc}2 & 4 \\ 8 & 2 \\ 10 & 12\end{array}\right]+\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]=\left[\begin{array}{cc}3 & 6 \\ 14 & 6 \\ 17 & 15\end{array}\right]$
$$\begin{aligned} \text{and}\quad (2 A+B)^{\prime} & =\left[\begin{array}{lll} 3 & 14 & 17 \\ 6 & 6 & 15 \end{array}\right] \\ \text{Also,}\quad 2 A^{\prime}+B^{\prime} & =2\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 1 & 6 \end{array}\right]+\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right] \\ & =\left[\begin{array}{lll} 3 & 14 & 17 \\ 6 & 6 & 15 \end{array}\right]=(2 A+B)^{\prime}\quad\text{Hence proved.} \end{aligned}$$
(ii) $(A-B)=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]=\left[\begin{array}{cc}0 & 0 \\ -2 & -3 \\ -2 & 3\end{array}\right]$
$$\begin{aligned} &\text { and }\\ &\begin{aligned} (A-B)^{\prime} & =\left[\begin{array}{llr} 0 & -2 & -2 \\ 0 & -3 & 3 \end{array}\right] \\ \text{Also,}\quad A^{\prime}-B^{\prime} & =\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 1 & 6 \end{array}\right]-\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right] \\ & =\left[\begin{array}{llr} 0 & -2 & -2 \\ 0 & -3 & 3 \end{array}\right] \\ & =(A-B)^{\prime}\quad\text{Hence proved.} \end{aligned} \end{aligned}$$
Show that $A^{\prime} A$ and $A A^{\prime}$ are both symmetric matrices for any matrix A.
$$\begin{aligned} \text{Let}\quad P & =A^{\prime} A \\ \therefore\quad P^{\prime} & =\left(A A^{\prime}\right)^{\prime} \\ & =A^{\prime}\left(A^{\prime}\right)^{\prime} \quad \left[\because\left(A B^{\prime}\right)^{\prime}=B^{\prime} A^{\prime}\right]\\ & =A^{\prime} A=P \end{aligned}$$
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.
Let $A$ and $B$ be square matrices of the order $3 \times 3$. Is $(A B)^2=A^2 B^2$ ? Give reasons.
$$\begin{aligned} &\text { Since, } A \text { and } B \text { are square matrices of order } 3 \times 3 \text {. }\\ &\therefore \quad A B^2=A B \cdot A B \end{aligned}$$
$$\begin{aligned} & =A B A B \\ & =A A B B \\ & =A^2 B^2 \end{aligned} \quad[\because A B=B A]$$
So, $A B^2=A^2 B^2$ is true when $A B=B A$.
Show that, if $A$ and $B$ are square matrices such that $A B=B A$, then $(A+B)^2=A^2+2 A B+B^2$
$$\begin{aligned} &\text { Since, } A \text { and } B \text { are square matrices such that } A B=B A \text {. }\\ &\begin{aligned} \therefore \quad(A+B)^2 & =(A+B) \cdot(A+B) \\ & =A^2+A B+B A+B^2 \\ & =A^2+A B+A B+B^2 \quad [\because A B=B A]\\ & =A^2+2 A B+B^2\quad \text{Hence proved.} \end{aligned} \end{aligned}$$