If and x2 = –1, then show
Question:

If and x2 = –1, then show that (A + B)2 = A2 + B2

Solution:

Given,

$A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ and $x^{2}=-1$

So, $(A+B)=\left[\begin{array}{cc}0 & -x+1 \\ x+1 & 0\end{array}\right]$

And,

And, $(A+B)^{2}=\left[\begin{array}{cc}0 & -x+1 \\ x+1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -x+1 \\ x+1 & 0\end{array}\right]=\left[\begin{array}{cc}1-x^{2} & 0 \\ 0 & 1-x^{2}\end{array}\right]$

And, $(A+B)^{2}=\left[\begin{array}{cc}0 & -x+1 \\ x+1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -x+1 \\ x+1 & 0\end{array}\right]=\left[\begin{array}{cc}1-x^{2} & 0 \\ 0 & 1-x^{2}\end{array}\right]$

Also, $A^{2}=A \cdot A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]=\left[\begin{array}{cc}-x^{2} & 0 \\ 0 & -x^{2}\end{array}\right]$

And,

$B^{2}=B \cdot B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Thus, $A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2} & 0 \\ 0 & -x^{2}\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1-x^{2} & 0 \\ 0 & 1-x^{2}\end{array}\right]$

Hence, from equations (i) and (ii), we have

$(A+B)^{2}=A^{2}+B^{2}$

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