Q97 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Mathematics
If \(A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4\end{array}\right]\) and \(B=A^3\), then \(B^{-1}=\)
- A\(\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -4\end{array}\right]\)
- B\(\left[\begin{array}{ccc}-27 & 0 & 0 \\ 0 & -125 & 0 \\ 0 & 0 & -64\end{array}\right]\)
- Correct answerC\(\left[\begin{array}{ccc}\frac{1}{27} & 0 & 0 \\ 0 & \frac{1}{125} & 0 \\ 0 & 0 & \frac{1}{64}\end{array}\right]\)
- D\(\left[\begin{array}{ccc}\frac{-1}{27} & 0 & 0 \\ 0 & \frac{-1}{125} & 0 \\ 0 & 0 & \frac{-1}{64}\end{array}\right]\)
Solution
The given matrix \(A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4\end{array}\right]\) and
\(\begin{aligned}
B & =A^3=\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right]\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right]\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right] \\
& =\left[\begin{array}{ccc}
9 & 0 & 0 \\
0 & 25 & 0 \\
0 & 0 & 16
\end{array}\right]\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{ccc}
27 & 0 & 0 \\
0 & 125 & 0 \\
0 & 0 & 64
\end{array}\right] \\
\therefore & \\
B^{-1} & =\frac{1}{27 \times 125 \times 64}\left[\begin{array}{ccc}
125 \times 64 & 0 & 0 \\
0 & 27 \times 64 & 0 \\
0 & 0 & 527 \times 125
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\frac{1}{27} & 0 & 0 \\
0 & \frac{1}{125} & 0 \\
0 & 0 & \frac{1}{64}
\end{array}\right]
\end{aligned}\)
Hence, option (c) is correct.
Hence, the correct answer is option: 'C'.