Question

Find the inverse of the given matrix
$\left[\begin{array}{ccc}1& 0& 0\\ 0& cos\alpha & sin\alpha \\ 0& sin\alpha & -cos\alpha \end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& cos\alpha & sin\alpha \\ 0& sin\alpha & -cos\alpha \end{array}\right]$
We have, $|A|=\left[\begin{array}{ccc}1& 0& 0\\ 0& cos\alpha & sin\alpha \\ 0& sin\alpha & -cos\alpha \end{array}\right]=1(-co{s}^2\alpha -si{n}^2\alpha )=-(co{s}^2\alpha +si{n}^2\alpha )=-1[\because co{s}^2\theta +si{n}^2\theta =1]$
Cofactos of A are
$A_{11}=-co{s}^2\alpha -si{n}^2\alpha =-1,A_{12}=-(0-0)=0,A_{13}=0-0=0A_{21}=-(0-0)=0,A_{22}=-cos\alpha -0=-cos\alpha A_{23}=-(sin\alpha -0)=-sin\alpha A_{31}=0-0=0,A_{32}=-(sin\alpha -0)=-sin\alpha ,A_{33}=cos\alpha -0=cos\alpha$

$\therefore adj\left(A\right)={\left[\begin{array}{ccc}-1& 0& 0\\ 0& -cos\alpha & -sin\alpha \\ 0& -sin\alpha & cos\alpha \end{array}\right]}^T=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -cos\alpha & -sin\alpha \\ 0& -sin\alpha & cos\alpha \end{array}\right]$

Now, $A^{-1}=\frac{1}{|A|}\left(adjA\right)$
$=\frac{1}{-1}\left[\begin{array}{ccc}-1& 0& 0\\ 0& cos\alpha & -sin\alpha \\ 0& -sin\alpha & cos\alpha \end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& cos\alpha & sin\alpha \\ 0& sin\alpha & -cos\alpha \end{array}\right]$