Browse Questions

# Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1&0&0 \\ 0&cos\alpha&sin\alpha \\ 0&sin\alpha&-cos\alpha \\ \end{bmatrix}$

Toolbox:
• (i)A matrix is said to be singular if $|A|$ =0.
• (ii)A matrix is said to be invertible only if $|A|\neq 0$.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
• (iv)The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$[A_{ij}]n\times n$
• where $A_{ij}$ is the cofactor of the element $[a_{ij}].$
Let A=$\begin{bmatrix}1 & 0& 0\\0 & cos \alpha& sin\alpha\\0 & sin\alpha & -cos \alpha\end{bmatrix}$
Expanding along $R_1$ we get,
$|A|=1(cos\alpha\times -cos \alpha-sin\alpha\times sin\alpha)-0+0$
$\;\;\;=-cos^2\alpha-sin^2\alpha=(sin^2\alpha+cos ^2\alpha).$
But $sin^2\alpha+cos^2\alpha=1.$
|A|=-1.
To find the adjoint A,let us find the minors and cofactors of the elements.
$M_{11}=\begin{bmatrix}cos\alpha & sin\alpha\\-sin\alpha & -cos\alpha\end{bmatrix}=-cos^2\alpha-sin^2\alpha$=-1M_{12}=\begin{bmatrix}0 & sin\alpha\\0 & -cos\alpha\end{bmatrix}=0-0=0M_{13}=\begin{bmatrix}0 & cos\alpha\\0 & sin\alpha\end{bmatrix}=0-0=0M_{21}=\begin{bmatrix}0 & 0\\sin\alpha & -cos\alpha\end{bmatrix}=0-0=0.M_{22}=\begin{bmatrix}1 & 0\\0 & -cos\alpha\end{bmatrix}=-cos\alpha-0=-cos\alphaM_{23}=\begin{bmatrix}1 & 0\\0 & sin\alpha\end{bmatrix}=sin\alphaM_{31}=\begin{bmatrix}0 & 0\\cos\alpha & sin\alpha\end{bmatrix}=0-0=0M_{32}=\begin{bmatrix}1 & 0\\0 & sin\alpha\end{bmatrix}=sin\alpha-0=sin\alphaM_{33}=\begin{bmatrix}1 & 0\\0 & cos\alpha\end{bmatrix}=cos\alpha-0=cos\alphaA_{11}=(-1)^{1+1}.(-1)=-1.A_{12}=(-1)^{1+2}.(0)=0.A_{13}=(-1)^{1+3}.(0)=0.A_{21}=(-1)^{2+1}.(0)=0.A_{22}=(-1)^{2+2}.(-cos\alpha)=-cos\alpha.A_{23}=(-1)^{2+3}\times sin\alpha=-sin\alpha.A_{31}=(-1)^{3+1}.(0)=0.A_{32}=(-1)^{3+2}.(sin\alpha)=-sin\alpha.A_{33}=(-1)^{3+3}.cos\alpha=cos\alpha.$Hence adj A=$\begin{bmatrix}A_{11} & A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} & A_{33}\end{bmatrix}\;\;\;\;\qquad\quad=\begin{bmatrix}-1 & 0 & 0\\0 & -cos\alpha & -sin\alpha\\0 & -sin\alpha & cos\alpha\end{bmatrix}$The inverse of the matrix is$A^{-1}=\frac{1}{|A|}adj \;A$|A|=-1. Hence$A^{-1}= -1\begin{bmatrix}-1 & 0 & 0\\0 & -cos\alpha & -sin\alpha\\0 & -sin\alpha & cos\alpha\end{bmatrix}A^{-1}= \begin{bmatrix}1 & 0 & 0\\0 & cos\alpha & sin\alpha\\0 & sin\alpha & -cos\alpha\end{bmatrix}\$