Q54 / 110 · VITEEE 2022 · Mathematics
If \(A=\left[\begin{array}{ll}3 & 2 \\ 0 & 1\end{array}\right]\), then \(\left(A^{-1}\right)^3\) is equal to
- Correct answerA\(\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]\)
- B\(\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \\ 0 & 27\end{array}\right]\)
- C\(\frac{1}{27}\left[\begin{array}{ll}1 & -26 \\ 0 & -27\end{array}\right]\)
- D\(\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \\ 0 & -27\end{array}\right]\)
Solution
\(|\mathrm{A}|=\left|\begin{array}{ll}
3 & 2 \\
0 & 1
\end{array}\right|=3-0=3 \neq 0 \quad \therefore \mathrm{A}^{-1} \text { exists }\)
Adj. \(A=\left[\begin{array}{cc}1 & 0 \\ -2 & 3\end{array}\right]^{\prime}=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]\)
\(\therefore \quad \mathrm{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]\)
\(\begin{aligned}
& \therefore\left(A^{-1}\right)^3=\frac{1}{27}\left[\begin{array}{cc}
1 & -2 \\
1 & 3
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right] \\
& =\frac{1}{27}\left[\begin{array}{cc}
1 & -8 \\
0 & 9
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]=\frac{1}{27}\left[\begin{array}{cc}
1 & -26 \\
0 & 27
\end{array}\right]
\end{aligned}\)
Hence, the correct answer is option: 'A'.