Question

Find the adjoint of the following matrix:

\[\begin{bmatrix}1 & \tan \alpha/2 \\ - \tan \alpha/2 & 1\end{bmatrix}\]
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.
\[\ D = \begin{bmatrix}1 & \tan\frac{\alpha}{2} \\ - \tan\frac{\alpha}{2} & 1\end{bmatrix}\]
\[adjD = \begin{bmatrix}1 & - \tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 1\end{bmatrix}\]
\[(adjD)D = \begin{bmatrix}1 + \tan^2 \frac{\alpha}{2} & 0 \\ 0 & 1 + \tan^2 \frac{\alpha}{2}\end{bmatrix}\]
\[\left| D \right| = 1 + \tan^2 \frac{\alpha}{2}\]
\[\left| D \right|I = \begin{bmatrix}1 + \tan^2 \frac{\alpha}{2} & 0 \\ 0 & 1 + \tan^2 \frac{\alpha}{2}\end{bmatrix}\]
\[D(adjD) = \begin{bmatrix}1 + \tan^2 \frac{\alpha}{2} & 0 \\ 0 & 1 + \tan^2 \frac{\alpha}{2}\end{bmatrix}\]
\[ \therefore (adjD)D = \left| D \right|I = D(adjD)\]
Hence verified.