Question

Given the matrices $A$ and $B$ as $A=\left[\begin{array}{cc}1& -1\\ 4& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -1\\ 2& -2\end{array}\right]$. The two matrices $X$ and $Y$ are such that $XA=B$ and $AY=B$. Then $3(X+Y)$ is equal to

• A. $\left[\begin{array}{cc}4& 1\\ -4& 2\end{array}\right]$
• B. $\left[\begin{array}{cc}4& -1\\ 4& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}-1& 4\\ 4& 2\end{array}\right]$
• D. $\left[\begin{array}{cc}1& -4\\ 4& 2\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}4& -1\\ 4& 2\end{array}\right]$

$A=\left[\begin{array}{cc}1& -1\\ 4& -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -1\\ 2& -2\end{array}\right]$

Here, $A$ is non-singular matrix but $B$ is singular.
Hence, only $A^{-1}$ exists.

Now, $XA=B$
$\Rightarrow X=B{A}^{-1}$
and $AY=B\Rightarrow Y=A^{-1}B$
$A^{-1}=\frac{1}{3}\left[\begin{array}{cc}-1& 1\\ -4& 1\end{array}\right]$
$\Rightarrow X=B{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}1& -1\\ 2& -2\end{array}\right]\left[\begin{array}{cc}-1& 1\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 0\end{array}\right]$

and $Y=A^{-1}B=\frac{1}{3}\left[\begin{array}{cc}-1& 1\\ -4& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& -2\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}1& -1\\ -2& 2\end{array}\right]$

$\therefore 3(X+Y)=\left[\begin{array}{cc}3& 0\\ 6& 0\end{array}\right]+\left[\begin{array}{cc}1& -1\\ -2& 2\end{array}\right]=\left[\begin{array}{cc}4& -1\\ 4& 2\end{array}\right]$