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 the matrix $3(X+Y)$

• A. $\left[\begin{array}{cc}-2& -1\\ -8& 2\end{array}\right]$
• B. $\left[\begin{array}{cc}4& -1\\ 4& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}4& 1\\ 4& -2\end{array}\right]$
• D. $\left[\begin{array}{cc}2& 1\\ 8& -2\end{array}\right]$

Answer 1

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

Here A is non singular but B is singular hence only $A^{-1}$ exists
Now $XA=B$
or $X=B{A}^{-1}$ (1)
And $AY=B$
or $Y=A^{-1}B$ (2)
Also $A^{-1}=\frac{1}{3}\left[\begin{array}{cc}-1& 1\\ -4& 1\end{array}\right]$
$\Rightarrow X=B{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}1& -1\\ 2& -2\end{array}\right]\left[\begin{array}{cc}-1& 1\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 0\end{array}\right]$
$\Rightarrow Y=A^{-1}B=\frac{1}{3}\left[\begin{array}{cc}-1& 1\\ -4& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& -2\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}1& -1\\ -2& 2\end{array}\right]$
$\Rightarrow 3(X+Y)=\left[\begin{array}{cc}3& 0\\ 6& 0\end{array}\right]+\left[\begin{array}{cc}1& -1\\ -2& 2\end{array}\right]=\left[\begin{array}{cc}4& -1\\ 4& 2\end{array}\right]$