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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the adjoint of the matrix: \[\begin{bmatrix} 1&-1&2 \\ 2&3&5 \\ -2&0&1 \end{bmatrix} \]

$\begin{array}{1 1} \begin{bmatrix}3 & 1 & -11\\-12 & 5 &1\\6 & 2 & 5\end{bmatrix} \\ \begin{bmatrix}3 & 1 & -11\\12 & 5 &-1\\6 & 2 & 5\end{bmatrix} \\ \begin{bmatrix}3 & 1 & -11\\-12 & 5 &-1\\6 & 2 & 5\end{bmatrix} \\ \begin{bmatrix}3 & 1 & 11\\-12 & 5 &-1\\6 & 2 & 5\end{bmatrix} \end{array} $

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Toolbox:
  • The adjoint of a square matrix is defined as the transpose of the matrix $[A_{ij}]_n\times n$ is the cofactor of the element $a_{ij}$
  • Adjoint of the matrix is denoted by adj A.
We know $A_{ij}=(-1)^{i+j}M_{ij}$
Here $M_{11}=\begin{vmatrix}3 & 5\\0 & 1\end{vmatrix}=3-0=3$
$M_{12}=\begin{vmatrix}2 & 5\\-2& 1\end{vmatrix}=2+10=12$
$M_{13}=\begin{vmatrix}2 & 3\\-2& 0\end{vmatrix}=0+6=6$
$M_{21}=\begin{vmatrix}-1 & 2\\0& 1\end{vmatrix}=-1+0=-1$
$M_{22}=\begin{vmatrix}1 & 2\\-2& 1\end{vmatrix}=1+4=5$
$M_{23}=\begin{vmatrix}1 & -1\\-2& 0\end{vmatrix}=0-2=-2$
$M_{31}=\begin{vmatrix}-1& 2\\3& 5\end{vmatrix}=-5-6=-11$
$M_{32}=\begin{vmatrix}1 & 2\\2& 5\end{vmatrix}=5-4=1$
$M_{33}=\begin{vmatrix}1 & -1\\2& 3\end{vmatrix}=3+2=5$
$A_{11}=(-1)^{1+1}.3=3$
$A_{12}=(-1)^{1+2}.12=-12$
$A_{13}=(-1)^{1+3}.6=6$
$A_{21}=(-1)^{2+1}.(-1)=1$
$A_{22}=(-1)^{2+2}.5=5$
$A_{23}=(-1)^{2+3}.(-2)=2$
$A_{31}=(-1)^{3+1}.(-11)=-11$
$A_{32}=(-1)^{3+2}.1=-1$
$A_{33}=(-1)^{3+3}.4=5$
Therefore $adj\; A=\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\;\;\;\qquad\qquad=\begin{bmatrix}3 & 1 & -11\\-12 & 5 &-1\\6 & 2 & 5\end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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