Question

If $\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, then $A=$

• A. $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$
• D. $-\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$

Let, $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Given, $\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}2a+c& 2b+d\\ 3a+2c& 3b+2d\end{array}\right]\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}-6a-3c+10b+5d& 4a+2c-6b-3d\\ -9a-6c+15b+10d& 6a+4c-9b-6d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Therefore,

$-6a-3c+10b+5d=1...\left(i\right)$

$4a+2c-6b-3d=0$

$\Rightarrow 4a+2c=6b+3d...\left(ii\right)$

$-9a-6c+15b+10d=0$

$\Rightarrow 9a+6c=15b+10d...\left(iii\right)$

$6a+4c-9b-6d=1...\left(iv\right)$

On solving the eqautions $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right)$, we get

$a=1,b=1,c=1$ $\&$ $d=0$

Therefore, The matrix $A$ becomes $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$.

Hence, the correct option is $\left(A\right)$