Question

Find the adjoint of the following matrix:
\[\begin{bmatrix}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\]

Verify that (adj A) A = |A| I = A (adj A) for the above matrix.

Answer 1

Given below is the square matrix. Here, we will interchange the diagonal elements and change the signs of the off-diagonal elements.
\[\ C = \begin{bmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}\]
\[adjC = \begin{bmatrix}\cos\alpha & - \sin\alpha \\ - \sin\alpha & \cos\alpha\end{bmatrix}\]
\[(adjC)C = \begin{bmatrix}\cos^2 \alpha - \sin^2 \alpha & 0 \\ 0 & \cos^2 \alpha - \sin^2 \alpha\end{bmatrix}\]
\[\left| C \right| = \cos^2 \alpha - \sin^2 \alpha\]
\[\left| C \right|I = \begin{bmatrix}\cos^2 \alpha - \sin^2 \alpha & 0 \\ 0 & \cos^2 \alpha - \sin^2 \alpha\end{bmatrix}\]
\[C(adjC) = \begin{bmatrix}\cos^2 \alpha - \sin^2 \alpha & 0 \\ 0 & \cos^2 \alpha - \sin^2 \alpha\end{bmatrix}\]
\[ \therefore (adjC)C = \left| C \right|I = C(adjC)\]
Hence verified.