Question

If $A$ is a matrix such that $\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)A\left(11\right)=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$ then $A=$

• A. $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$
• B. $\left(21\right)$
• C. $\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$
• D. $\left(\begin{array}{c}2\\ -3\end{array}\right)$

Answer 1

The correct option is D $\left(\begin{array}{c}2\\ -3\end{array}\right)$

$\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)A\left(\begin{array}{cc}1& 1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$

To maintain the rules of matrix multiplication, $A$ has to be of the order $2\times 1$

$\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\left(\begin{array}{cc}1& 1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}2x+y\\ 3x+2y\end{array}\right)\left(\begin{array}{cc}1& 1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$

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

$\Rightarrow 2x+y=1$ and $3x+2y=0$

$\Rightarrow 4x+2y-(3x+2y)=2-0$

$\Rightarrow x=2$ and $y=-3$