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Algebra of Matrices — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
If $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$ then show that $(A + B)^2 = A^2 + B^2.$
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Answer

Given: $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$
To show: $(A + B)^2 = A^2 + B^2​​​​​​​$​​​​​​​
L.H.S
$\text{A}+\text{B}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}+\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0+0&-\text{x}+1\\\text{x}+1&0+0\end{bmatrix}$
$=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$(\text{A}+\text{B})^2=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$=\begin{bmatrix}0+(1-\text{x})(1+\text{x})&0+0\\0+0&(\text{x}+1)(1-\text{x})\end{bmatrix}$
$=\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ \dots(1)$
R.H.S
$ \text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}$
$ \text{A}^2=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}$
$=\begin{bmatrix}0-\text{x}^2&0+0\\0+0&-\text{x}^2+0\end{bmatrix}$
$=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}\ \dots(2)$
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\text{B}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0+1&0+0\\0+0&1+0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(3)$
Adding (2) and (3), we get
$\text{A}^2+\text{B}^2=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ =\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ \dots(4)$
Comparing (1) and (4), we get
$(A + B)^2 = A^2 + B^2​​​​​​​$

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