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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the inverse of the matrix (if it exists): \[ \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \\ \end{bmatrix} \]

\begin{array}{1 1} A^{-1}= \begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix} \\ A^{-1}= \begin{bmatrix}2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix} \\ A^{-1}= \begin{bmatrix}-2 & 0 & -1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix} \\ A^{-1}= \begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & 2\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=$\begin{bmatrix}1 & -1& 2\\0 & 2& -3\\3 & -2 & 4\end{bmatrix}$
Expanding along $R_1$ we get,
$|A|=1(2\times 4-(-3\times -2))-(-1)(0\times 4-(-3\times 3)+2(0\times -2-(2\times 3)$
$\;\;\;=1[(8-6)+(0+9)+2(0-6)]=[2+9-12]=-1.$
Now to find the adjoint of A,
let us find the minors and cofactors of the elements.
$M_{11}=\begin{bmatrix}2 & -3\\-2 & 4\end{bmatrix}=8-6=2$
$M_{12}=\begin{bmatrix}0 & -3\\3 & 4\end{bmatrix}=0+9=9$
$M_{13}=\begin{bmatrix}0 & 2\\3 & -2\end{bmatrix}=0-6=-6$
$M_{21}=\begin{bmatrix}-1 & 2\\-2 & 4\end{bmatrix}=-4+4=0.$
$M_{22}=\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}=4-6=-2$
$M_{23}=\begin{bmatrix}1 & -1\\3 & -2\end{bmatrix}=-2+3=1$
$M_{31}=\begin{bmatrix}-1 & 2\\2 & -3\end{bmatrix}=3-4=-1$
$M_{32}=\begin{bmatrix}1 & 2\\0 & -3\end{bmatrix}=-3-0=-3$
$M_{33}=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}=2-0=2$
$A_{11}=(-1)^{1+1}.(2)=2.$
$A_{12}=(-1)^{1+2}.(9)=-9.$
$A_{13}=(-1)^{1+3}.(-6)=-6.$
$A_{21}=(-1)^{2+1}.(0)=0.$
$A_{22}=(-1)^{2+2}.(-2)=-2.$
$A_{23}=(-1)^{2+3}.1=-1.$
$A_{31}=(-1)^{3+1}.(-1)=-1.$
$A_{32}=(-1)^{3+2}.(-3)=3.$
$A_{33}=(-1)^{3+3}.2=2.$
Hence adj A=$\begin{bmatrix}A_{11} & A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} & A_{33}\end{bmatrix}$
$\;\;\;\;\qquad\quad=\begin{bmatrix}2 & 0 & -1\\-9 & -2 & 3\\-6 & -1 & 2\end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
|A|= -1.
Hence $A^{-1}= -1\begin{bmatrix}2 & 0 & -1\\-9 &- 2 & 3\\-6 & -1 & 2\end{bmatrix}$
$A^{-1}= \begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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