Q58 / 110 · VITEEE 2024 · Mathematics
If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$, then $\left(B B^T A\right)^5$ is equal to
- A$\left[\begin{array}{cc}2+\sqrt{3} & 1 \\ -1 & 2-\sqrt{3}\end{array}\right]$
- B$\frac{1}{2}\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]$
- Correct answerC$\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]$
- D$\left[\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right]$
Solution
$\begin{aligned}
& \text { } \because B=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{-1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right] \\
& \therefore \quad B^T=\left[\begin{array}{ll}
\frac{\sqrt{3}}{2} & \frac{-1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right] \\
& \text { and } B B^T=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
-\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right]\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{-1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
\frac{3}{4}+\frac{1}{4} & \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\
\frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4}
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& \text { and } A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \\
& \therefore\left(B B^T A\right)\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=A \\
& \text { and } \quad\left(B B^T A\right)^2=A^2=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
& \text { Hence, }\left(B B^T A\right)^5=\left[\begin{array}{ll}
1 & 5 \\
0 & 1
\end{array}\right]
\end{aligned}$
Hence, the correct answer is option: 'C'.