Question

If  \[A = \begin{pmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix},\] ,find adj·A and verify that A(adj·A) = (adj·A)A = |A| I₃.

Answer 1

Consider the matrix, 

\[A = \begin{pmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix},\] \[\text { adj }\left( A \right) = C^T\] Where, C is cofactor matrix.

\[C = \begin{pmatrix}\cos \alpha & - \sin\alpha & 0 \\ \sin \alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}, \]

\[\text { Adj }\left( A \right) = C^T = \begin{pmatrix}cos\alpha & sin\alpha & 0 \\ - sin\alpha & cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}\] 

Now,

\[A . \text { Adj }\left( A \right) = \begin{pmatrix}cos\alpha & - sin\alpha & 0 \\ sin\alpha & cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}cos\alpha & sin\alpha & 0 \\ - sin\alpha & cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}\]

\[ = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} . . . . . (1)\]

\[\text { Adj }\left( A \right) . A = \begin{pmatrix}cos\alpha & sin\alpha & 0 \\ - sin\alpha & cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}cos\alpha & - sin\alpha & 0 \\ sin\alpha & cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}\]

\[ = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} . . . . . (2)\]

\[\left| A \right| = \begin{vmatrix}cos\alpha & - sin\alpha & 0 \\ sin\alpha & cos\alpha & 0 \\ 0 & 0 & 1\end{vmatrix}\]

\[ = cos\alpha\left( cos\alpha - 0 \right) + sin\alpha\left( sin\alpha - 0 \right) + 0\]

\[ = 1 . . . . . (3)\]

From equation (1), (2) and (3), it is obtained that A(adj·A) = (adj·A)A = |A| I₃.