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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the inverse of the matrix (if it exists): \[ \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \\ \end{bmatrix} \]

\begin{array}{1 1}3\begin{bmatrix}-3 & 0 & 0\\3 & -1 & 0\\-9 & -2 & 3\end{bmatrix} \\ -3\begin{bmatrix}-3 & 0 & 0\\3 & -1 & 0\\-9 & -2 & 3\end{bmatrix} \\\frac{1}{3}\begin{bmatrix}-3 & 0 & 0\\3 & -1 & 0\\-9 & -2 & 3\end{bmatrix} \\ \frac{1}{-3}\begin{bmatrix}-3 & 0 & 0\\3 & -1 & 0\\-9 & -2 & 3\end{bmatrix}\end{array} $

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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=$\begin{bmatrix}1 & 0& 0\\3 & 3& 0\\5 & 2 & -1\end{bmatrix}$
Expanding along $R_1$ we get,
$|A|=1(-3-0)-0+0=-3.$
Since |A|$\neq 0$ inverse exists,
Now let us find the minors and cofactors of the elements.
$M_{11}=\begin{bmatrix}3 & 0\\2 & -1\end{bmatrix}=-3-0=-3$
$M_{12}=\begin{bmatrix}3 & 0\\5 & -1\end{bmatrix}=-3-0=-3$
$M_{13}=\begin{bmatrix}3 & 3\\5 & 2\end{bmatrix}=6-15=-9$
$M_{21}=\begin{bmatrix}0 & 0\\2 & -1\end{bmatrix}=0-0=0$
$M_{22}=\begin{bmatrix}1 & 0\\5 & -1\end{bmatrix}=-1-0=-1$
$M_{23}=\begin{bmatrix}1 & 0\\5 & 2\end{bmatrix}=2-0=2$
$M_{31}=\begin{bmatrix}0 & 0\\3 & 0\end{bmatrix}=0-0=0$
$M_{32}=\begin{bmatrix}1 & 0\\3 & 0\end{bmatrix}=0-0=0$
$M_{33}=\begin{bmatrix}1 & 0\\3 & 3\end{bmatrix}=3-0=3$
$A_{11}=(-1)^{1+1}.(-3)=-3.$
$A_{12}=(-1)^{1+2}.(-3)=3.$
$A_{13}=(-1)^{1+3}.(-9)=-9.$
$A_{21}=(-1)^{2+1}.0=0.$
$A_{22}=(-1)^{2+2}.(-1)=-1.$
$A_{23}=(-1)^{2+3}.2=-2.$
$A_{31}=(-1)^{3+1}.0=0.$
$A_{32}=(-1)^{3+2}.0=0.$
$A_{33}=(-1)^{3+3}.3=3.$
Therefore adj A=$\begin{bmatrix}A_{11} & A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} & A_{33}\end{bmatrix}$
$\;\;\;\;\qquad\quad=\begin{bmatrix}-3 & 0 & 0\\3 & -1 & 0\\-9 & -2 & 3\end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
|A|= -3
Therefore inverse =$\frac{1}{-3}\begin{bmatrix}-3 & 0 & 0\\3 & -1 & 0\\-9 & -2 & 3\end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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