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MATRICES — Maths STD 12 Science — Question

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

We have, $\text{A}=\begin{bmatrix}0&1\\1&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$\therefore\ (\text{A}+\text{B})=\begin{bmatrix}0+0&1-1\\1+1&1+0\end{bmatrix}$
$=\begin{bmatrix}0&0\\2&1\end{bmatrix}$
and $(\text{A}-\text{B})=\begin{bmatrix}0-0&1+1\\1-1&1-0\end{bmatrix}$
$=\begin{bmatrix}0&2\\0&1\end{bmatrix}$
$(\text{A}+\text{B}).(\text{A}-\text{B})=\begin{bmatrix}0&0\\2&1\end{bmatrix}\begin{bmatrix}0&2\\0&1\end{bmatrix}$
$=\begin{bmatrix}0+0&0+0\\0+0&4+1\end{bmatrix}=\begin{bmatrix}0&0\\0&5\end{bmatrix}\ ....(\text{i})$
Also, $\text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}0&1\\1&1\end{bmatrix}.\begin{bmatrix}0&1\\1&1\end{bmatrix}$
$=\begin{bmatrix}0+1&0+1\\0+1&1+1\end{bmatrix}=\begin{bmatrix}1&1\\1&2\end{bmatrix}$
and $\text{B}^2=\text{B}.\text{B}$
$=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0-1&0+0\\0+0&-1+0\end{bmatrix}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$
$\therefore\ \text{A}^2-\text{B}^2=\begin{bmatrix}1&1\\1&2\end{bmatrix}-\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$
$=\begin{bmatrix}2&1\\1&3\end{bmatrix}\ ....(\text{ii})$
From (i) and (ii), $(\text{A}+\text{B})(\text{A}-\text{B})\neq\text{A}^2-\text{B}^2$

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