Question

Let $A=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right],B=\left[\begin{array}{cc}x& 1\\ y& -1\end{array}\right]$, if $(A+B{)}^2=A^2+B^2$, find $(x,y)$ is equal to

• A. $(4,1)$
• B. $(1,4)$
• C. $(-4,-1)$
• D. $(-1,-4)$

Answer 1

The correct option is B $(1,4)$

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

$\therefore AB=-BA$

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

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

Comparing, we get

$x+1=2,y-1=3$

$\therefore x=1,y=4$

$(x,y)=(1,4)$