If $A\left[\begin{array}{cc}3 & -4 \\ 1 & 1 \\ 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$, then verify $(B A)^2 \neq B^2 A^2$.
We have, $$ A=\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]_{3 \times 2} \text { and } B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]_{2 \times 3}$$
$\therefore\quad B A=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}\left[\begin{array}{cc}3 & -4 \\ 1 & 1 \\ 2 & 0\end{array}\right]_{3 \times 2}$
$=\left[\begin{array}{ll}6+1+4 & -8+1+0 \\ 3+2+8 & -4+2+0\end{array}\right]=\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right]$
$$\begin{aligned} \text{and}\quad (B A) \cdot(B A) & =\left[\begin{array}{ll} 11 & -7 \\ 13 & -2 \end{array}\right]\left[\begin{array}{cc} 11 & -7 \\ 13 & -2 \end{array}\right] \\ \Rightarrow\quad (B A)^2 & =\left[\begin{array}{cc} 121-91 & -77+14 \\ 143-26 & -91+4 \end{array}\right]=\left[\begin{array}{cc} 30 & -63 \\ 117 & -87 \end{array}\right] \quad\text{... (i)}\\ \text{Also,}\quad B^2 & =B \cdot B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]_{2 \times 3}\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]_{2 \times 3} \end{aligned}$$
So, $B^2$ is not possible, since the $B$ is not a square matrix.
Hence, $(B A)^2 \neq B^2 A^2$.
If possible, find the value of $B A$ and $A B$, where $$ A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right] \text {. }$$
We have, $A=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}$ and $B=\left[\begin{array}{ll}4 & 1 \\ 2 & 3 \\ 1 & 2\end{array}\right]_{3 \times 2}$
So, $A B$ and $B A$ both are possible.
[since, in both $A \cdot B$ and $B \cdot A$, the number of columns of first is equal to the number of rows of second.]
$$\begin{aligned} \therefore\quad A B & =\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]_{2 \times 3} \cdot\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]_{3 \times 2} \\ & =\left[\begin{array}{ll} 8+2+2 & 2+3+4 \\ 4+4+4 & 1+6+8 \end{array}\right]=\left[\begin{array}{cc} 12 & 9 \\ 12 & 15 \end{array}\right] \end{aligned}$$
$$\begin{aligned} \text{and}\quad B A & =\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]_{3 \times 2}\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]_{2 \times 3} \\ & =\left[\begin{array}{ccc} 4 \times 2+1 & 4+2 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8 \end{array}\right]=\left[\begin{array}{lll} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{array}\right] \end{aligned}$$
Show by an example that for $A \neq 0, B \neq 0$ and $A B=0$.
$$\begin{aligned} \text{Let}\quad A & =\left[\begin{array}{cc} 0 & -4 \\ 0 & 2 \end{array}\right] \neq 0 \text { and } B=\left[\begin{array}{ll} 3 & 5 \\ 0 & 0 \end{array}\right] \neq 0 \\ \therefore\quad A B & =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{aligned}$$
Hence proved.
Given, $A=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]$. is $(A B)^{\prime}=B^{\prime} A^{\prime}$ ?
We have, $A=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right]_{2 \times 3}$ and $B=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]_{3 \times 2}$
$\therefore\quad A B=\left[\begin{array}{cc}2+8+0 & 8+32+0 \\ 3+18+6 & 12+72+18\end{array}\right]=\left[\begin{array}{cc}10 & 40 \\ 27 & 102\end{array}\right]$
$\text{and}\quad (A B)^{\prime}=\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right]\quad\text{..... (i)}$
$\text{Also,}\quad B^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right]_{2 \times 3}$ and $A^{\prime}=\left[\begin{array}{ll}2 & 3 \\ 4 & 9 \\ 0 & 6\end{array}\right]_{3 \times 2}$
$\therefore\quad B^{\prime} A^{\prime}=\left[\begin{array}{cc}2+8+0 & 3+18+6 \\ 8+32+0 & 12+72+18\end{array}\right]=\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right]\quad\text{.... (ii)}$
Thus, we see that, $(A B)^{\prime}=B^{\prime} A^{\prime}$ [using Eqs. (i) and (ii)]
Solve for $x$ and $y, x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=0$.
$$\begin{aligned} \text{We have,}\quad x\left[\begin{array}{l} 2 \\ 1 \end{array}\right]+y\left[\begin{array}{l} 3 \\ 5 \end{array}\right]+\left[\begin{array}{c} -8 \\ -11 \end{array}\right] & =0 \\ \Rightarrow\quad{\left[\begin{array}{c} 2 x \\ x \end{array}\right]+\left[\begin{array}{l} 3 \cdot y \\ 5 \cdot y \end{array}\right]+\left[\begin{array}{c} -8 \\ -11 \end{array}\right] } & =0 \\ \Rightarrow\quad{\left[\begin{array}{cc} 2 x & 3 y \\ x & -8 \\ x & -11 \end{array}\right] } & =\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \\ \therefore\quad 2 x+3 y-8 & =0 \\ \Rightarrow\quad 4 x+6 y & =16 \quad\text{.... (i)}\\ \text{and}\quad x+5 y-11 & =0 \\ \Rightarrow\quad 4 x+20 y & =44\quad\text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { On subtracting Eq. (i) from Eq. (ii), we get }\\ &\begin{array}{lc} & 14 y=28 \Rightarrow y=2 \\ \therefore & 2 x+3 \times 2-8=0 \\ \Rightarrow & 2 x=2 \Rightarrow x=1 \\ \therefore & x=1 \text { and } y=2 \end{array} \end{aligned}$$