Question

Find co-factors of the matrix,

$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right]$

Here, $|A|=-{\cos }^2\alpha -{\sin }^2\alpha$

$\Rightarrow |A|=-1$ ($\because {\sin }^2\alpha +{\cos }^2\alpha =1$)

Since, $|A|\ne 0$

Hence, $A^{-1}$ exists.

We have $A^{-1}=\frac{adjA}{|A|}$

and $adjA=C^T$

So, we will find the co-factors of each element of $A$.

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right|$

$\Rightarrow C_{11}=-({\sin }^2\alpha +{\cos }^2\alpha )=-1$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}0& \sin \alpha \\ 0& -\cos \alpha \end{array}\right|$

$\Rightarrow C_{12}=-(0-0)=0$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}0& \cos \alpha \\ 0& \sin \alpha \end{array}\right|$

$\Rightarrow C_{13}=0-0=0$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}0& 0\\ \sin \alpha & -\cos \alpha \end{array}\right|$

$\Rightarrow C_{21}=-(0-0)=0$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 0\\ 0& -\cos \alpha \end{array}\right|$

$\Rightarrow C_{22}=(-\cos \alpha -0)=-\cos \alpha$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 0\\ 0& \sin \alpha \end{array}\right|$

$\Rightarrow C_{23}=-(\sin \alpha -0)=-\sin \alpha$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}0& 0\\ \cos \alpha & \sin \alpha \end{array}\right|$

$\Rightarrow C_{31}=(0-0)=0$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 0\\ 0& \sin \alpha \end{array}\right|$

$\Rightarrow C_{32}=-(\sin \alpha -0)=-\sin \alpha$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 0\\ 0& \cos \alpha \end{array}\right|$

$\Rightarrow C_{33}=(\cos \alpha -0)=\cos \alpha$

Hence, the co-factor matrix of A is

$C=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -\cos \alpha & -\sin \alpha \\ 0& -\sin \alpha & \cos \alpha \end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -\cos \alpha & -\sin \alpha \\ 0& -\sin \alpha & \cos \alpha \end{array}\right]$

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

$\Rightarrow A^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right]$