Q62 / 120 · VITEEE 2011 · Mathematics
If $\mathrm{I}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right], \mathrm{J}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then $\mathrm{B}$ is equal to
- Correct answerA$\quad I \cos \theta+J \sin \theta$
- B$I \sin \theta+J \cos \theta$
- CI $\cos \theta-\mathrm{J} \sin \theta$
- D$-\mathrm{I} \cos \theta+\mathrm{J} \sin \theta$
Solution
$$
\begin{aligned}\quad \mathrm{B} &=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
&=\left[\begin{array}{cc}
\cos \theta & 0 \\
0 & \cos \theta
\end{array}\right]+\left[\begin{array}{cc}
0 & \sin \theta \\
-\sin \theta & 0
\end{array}\right] \\
&=\cos \theta\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right] \\
&=\mathrm{I} \cos \theta+\mathrm{J} \sin \theta
\end{aligned}
$$
Hence, the correct answer is option: 'A'.