Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (D)

${\left(A+B\right)}^2=A^2+B^2\Rightarrow AB+BA=0$

$\Rightarrow \left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]\left[\begin{array}{cc}x& 1\\ y& -1\end{array}\right]+\left[\begin{array}{cc}x& 1\\ y& -1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]=0$

$\Rightarrow \left[\begin{array}{cc}x-y& 2\\ 2x-y& 3\end{array}\right]+\left[\begin{array}{cc}x+2& -x-1\\ y-2& -y+1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow 2x-y+2=\mathrm{0....}\left(i\right),-x+1=\mathrm{0....}\left(ii\right)$

$2x-2=\mathrm{0....}\left(iii\right)and-y+4=0\backslash ,\backslash ,....\left(iv\right)$

From $\left(ii\right),x=1$ and from $\left(iv\right),y=4$

Now, $x+y=1+4=5$