Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Find the inverse of each of the matrices (if it exists).

Question:

Find the inverse of each of the matrices (if it exists).

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

Solution:

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

We have,

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

Now,

$A_{11}=-\cos ^{2} \alpha-\sin ^{2} \alpha=-1, A_{12}=0, A_{13}=0$

$A_{21}=0, A_{22}=-\cos \alpha, A_{23}=-\sin \alpha$

$A_{31}=0, A_{32}=-\sin \alpha, A_{33}=\cos \alpha$

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

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