Q115 / 180 · KCET 2020 · Mathematics
If $\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, then the matrix $a$ is
- A$\left(\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right)$
- Correct answerB$\left(\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right)$
- C$\left(\begin{array}{cc}-2 & 1 \\ 3 & -2\end{array}\right)$
- D$\left(\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right)$
Solution
We have,
$\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Let
$B=\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right]$
$\therefore \quad B A=I$
$A=B^{-1} I$
$A=B^{-1}$
$\therefore \quad B^{-1}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$
Hence, $A=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$
Hence, the correct answer is option: 'B'.