Q47 / 160 · WBJEE 2010 · Mathematics
If $\mathrm{A}=\left[\begin{array}{rr}1 & 2 \\ -4 & -1\end{array}\right]$ then $\mathrm{A}^{-1}$ is
- Correct answerA$\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & 1\end{array}\right]$
- B$\frac{1}{7}\left[\begin{array}{rr}1 & 2 \\ -4 & -1\end{array}\right]$
- C$\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & -1\end{array}\right]$
- DDoes not exist
Solution
Hints: $|\mathrm{A}|=-1+8=7$
$$
\begin{aligned}
& \operatorname{adj}(\mathrm{A})=\left[\begin{array}{rr}
+(-1) & -(2) \\
-(-4) & +(1)
\end{array}\right]=\left[\begin{array}{rr}
-1 & -2 \\
4 & 1
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{7}\left[\begin{array}{rr}
-1 & -2 \\
4 & 1
\end{array}\right] \quad \operatorname{Both}(\mathrm{A} \text { and } C)
\end{aligned}
$$
Hence, the correct answer is option: 'A'.