Q74 / 180 · COMEDK 2023 (Memory Based) · Mathematics
If $A=\left[\begin{array}{ll}2 & 2 \\ 3 & 4\end{array}\right]$, then $A^{-1}$ equals to
- A$\left[\begin{array}{cc}2 & 1 \\ -3 / 2 & -1\end{array}\right]$
- Correct answerB$\left[\begin{array}{cc}2 & -1 \\ -3 / 2 & 1\end{array}\right]$
- C$\left[\begin{array}{cc}-2 & 1 \\ 3 / 2 & -1\end{array}\right]$
- D$\left[\begin{array}{cc}-2 & -1 \\ 3 / 2 & 1\end{array}\right]$
Solution
$$
\begin{aligned}
& \text { Given, } A=\left[\begin{array}{ll}
2 & 2 \\
3 & 4
\end{array}\right] \\
& \therefore \quad|A|=\left|\begin{array}{ll}
2 & 2 \\
3 & 4
\end{array}\right|=2 \times 4-3 \times 2=8-6=2
\end{aligned}
$$
Now, $A_{11}=4, A_{12}=-3, A_{21}=-2$ and $A_{22}=2$
$$
\begin{aligned}
\therefore \quad a d j_A & =\left[\begin{array}{cc}
4 & -3 \\
-2 & 2
\end{array}\right]^T=\left[\begin{array}{cc}
4 & -2 \\
-3 & 2
\end{array}\right] \\
A^{-1} & =\frac{1}{|A|} \operatorname{adj} A=\frac{1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 & -1 \\
-3 / 2 & 1
\end{array}\right]
\end{aligned}
$$
Hence, the correct answer is option: 'B'.