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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
If $\text{A}=\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix},$ then verify $(\text{BA})^2\neq\text{B}^2\text{A}^2.$
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Answer

We have, $\text{A}=\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}_{3\times2}$ and $\text{B}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}$
$\therefore\ \text{BA}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}_{3\times2}$
$=\begin{bmatrix}6+1+4&-8+1+0\\3+2+8&-4+2+0\end{bmatrix}=\begin{bmatrix}11&-7\\13&-2\end{bmatrix}$
And $(\text{BA}).(\text{BA})=\begin{bmatrix}11&-7\\13&-2\end{bmatrix}.\begin{bmatrix}11&-7\\13&-2\end{bmatrix}$
$\Rightarrow\ (\text{BA})^2=\begin{bmatrix}121-91&-77+14\\143-26&-91+4\end{bmatrix}$
$=\begin{bmatrix}30&-63\\117&-87\end{bmatrix}\ ....(\text{i})$ Also, $\text{B}^2=\text{B}.\text{B}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}.\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}$
So $B^2$ is not possible, since the B is not a square matrix.
Hence, $(\text{BA})^2\neq\text{B}^2\text{A}^2.$

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