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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the inverse of the matrix (if it exists): \[ \begin{bmatrix} 1&0&0 \\ 0&cos\alpha&sin\alpha \\ 0&sin\alpha&-cos\alpha \\ \end{bmatrix} \]

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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 $[a_{ij}].$
  • Adjoint is denoted by adj A.
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.
$M_{11}=\begin{bmatrix}cos\alpha & sin\alpha\\-sin\alpha & -cos\alpha\end{bmatrix}=-cos^2\alpha-sin^2\alpha$=-1$
$M_{12}=\begin{bmatrix}0 & sin\alpha\\0 & -cos\alpha\end{bmatrix}=0-0=0$
$M_{13}=\begin{bmatrix}0 & cos\alpha\\0 & sin\alpha\end{bmatrix}=0-0=0$
$M_{21}=\begin{bmatrix}0 & 0\\sin\alpha & -cos\alpha\end{bmatrix}=0-0=0.$
$M_{22}=\begin{bmatrix}1 & 0\\0 & -cos\alpha\end{bmatrix}=-cos\alpha-0=-cos\alpha$
$M_{23}=\begin{bmatrix}1 & 0\\0 & sin\alpha\end{bmatrix}=sin\alpha$
$M_{31}=\begin{bmatrix}0 & 0\\cos\alpha & sin\alpha\end{bmatrix}=0-0=0$
$M_{32}=\begin{bmatrix}1 & 0\\0 & sin\alpha\end{bmatrix}=sin\alpha-0=sin\alpha$
$M_{33}=\begin{bmatrix}1 & 0\\0 & cos\alpha\end{bmatrix}=cos\alpha-0=cos\alpha$
$A_{11}=(-1)^{1+1}.(-1)=-1.$
$A_{12}=(-1)^{1+2}.(0)=0.$
$A_{13}=(-1)^{1+3}.(0)=0.$
$A_{21}=(-1)^{2+1}.(0)=0.$
$A_{22}=(-1)^{2+2}.(-cos\alpha)=-cos\alpha.$
$A_{23}=(-1)^{2+3}\times sin\alpha=-sin\alpha.$
$A_{31}=(-1)^{3+1}.(0)=0.$
$A_{32}=(-1)^{3+2}.(sin\alpha)=-sin\alpha.$
$A_{33}=(-1)^{3+3}.cos\alpha=cos\alpha.$
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 $A^{-1}= -1\begin{bmatrix}-1 & 0 & 0\\0 & -cos\alpha & -sin\alpha\\0 & -sin\alpha & cos\alpha\end{bmatrix}$
$A^{-1}= \begin{bmatrix}1 & 0 & 0\\0 & cos\alpha & sin\alpha\\0 & sin\alpha & -cos\alpha\end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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