Question

Let $A=\left[\begin{array}{cc}x+y& y\\ 2x& x-y\end{array}\right],B=\left[\begin{array}{cc}2& -1\end{array}\right]$ and $C=\left[\begin{array}{cc}3& 2\end{array}\right].$ If $AB=C$, then $A^2$ is equal to

• A. $\left[\begin{array}{cc}6& -10\\ 4& 26\end{array}\right]$
• B. $\left[\begin{array}{cc}-10& 5\\ 4& 24\end{array}\right]$
• C. $\left[\begin{array}{cc}-5& -6\\ -4& -20\end{array}\right]$
• D. None of these.

Answer 1

The correct option is D None of these.

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

Since,

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

The multiplication of $A$ and $B$ can't be done because it doesn't satisfy the conditions of matrix multipication