If A= 0&-x &0 ,B= 0&1 1&0 and x^2 = -1 then show that (A + B)^2 = A^2 + B^2.

Given: A= 0&-x &0 ,B= 0&1 1&0 and x^2 = -1 To show: (A + B)^2 = A^2 + B^2 L.H.S A+B= 0&-x &0 + 0&1 1&0 = 0+0&-x+1 +1&0+0 = 0&-x+1 +1&0 (A+B)^2= 0&-x+1 +1&0 0&-x+1 +1&0 = 0+(1-x)(1+x)&0+0 0+0&(x+1)(1-x) = 1-x^2&0 0&1-x^2 (1) R.H.S A= 0&-x &0 A^2= 0&-x &0 0&-x &0 = 0-x^2&0+0 0+0&-x^2+0 = -x^2&0 0&-x^2 (2) B= 0&1 1&0 B^2= 0&1 1&0 0&1 1&0 = 0+1&0+0 0+0&1+0 = 1&0 0&1 (3) Adding (2) and (3), we get A^2+B^2= -x^2&0 0&-x^2 + 1&0 0&1 = 1-x^2&0 0&1-x^2 (4) Comparing (1) and (4), we get (A + B)^2 = A^2 + B^2

If A= 0&-x &0 ,B= 0&1 1&0 and x^2 = -1 then show that (A + B)^2 =…

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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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