Solve this

Question:

If $A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ and $x^{2}=-1$, then show that $(A+B)^{2}=A^{2}+B^{2}$.

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 $x^{2}=-1$

To show: $(A+B)^{2}=A^{2}+B^{2}$

LHS:

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

$=\left[\begin{array}{cc}0+0 & -x+1 \\ x+1 & 0+0\end{array}\right]$

$=\left[\begin{array}{cc}0 & -x+1 \\ x+1 & 0\end{array}\right]$

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

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

$=\left[\begin{array}{cc}1-x^{2} & 0 \\ 0 & 1-x^{2}\end{array}\right]$          ...(1)

RHS:

$A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]$

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

$=\left[\begin{array}{cc}0-x^{2} & 0+0 \\ 0+0 & -x^{2}+0\end{array}\right]$         ...(2)

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

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

$=\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$              ...(3)

Adding $(2)$ and $(3)$, we get

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

$=\left[\begin{array}{cc}1-x^{2} & 0 \\ 0 & 1-x^{2}\end{array}\right]$  ...(4)

Comparing (1) and (4), we get

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

 

 

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