Question

If $A=\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}$ and A adjA $=\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}$, then k is equal to
A. 0
B. 1
C. $\sin \alpha \cos \alpha$
D. $\cos 2 \alpha$

Answer 1

Hint: We are given a matrix, $A=\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}$. We have to find the value of k in the product of A and adjoint matrix of A.

Complete step-by-step solution: We have, $A=\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}$. Now to find the adjoint matrix, interchange the diagonal elements and multiply the off-diagonal elements by a negative sign.
Therefore, the adjoint matrix of A, adj A=$\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}$.
So, A (adj A)=$\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}$
$\begin{bmatrix}k & 0 \\0 & k\end{bmatrix}=\begin{bmatrix}\cos \alpha \times \cos \alpha+\sin \alpha \times \sin \alpha & \cos \alpha \times(-\sin \alpha)+\sin \alpha \times \cos \alpha \\
(-\sin \alpha) \times \cos \alpha+\cos \alpha \times \sin \alpha & (-\sin \alpha) \times(-\sin \alpha)+\cos \alpha \times \cos \alpha
\end{bmatrix}$
$\begin{bmatrix}k & 0 \\0 & k\end{bmatrix}=\begin{bmatrix}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\cos \alpha \sin \alpha+\cos \alpha \sin \alpha \\
-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha
\end{bmatrix}$
$\begin{bmatrix}k & 0 \\0 & k\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1
\end{bmatrix}$, since we have $\cos ^{2} \alpha+\sin ^{2} \alpha=1$
Therefore, the value of k is 1 .
The correct answer is Option B.

Note: Recall the formula (adj A) A=A(adj A)=|A| I, where I is the identity matrix. If we look at the question we need to calculate |A|I, so it is enough to calculate the determinant of the given matrix instead of multiplying A and adj A.