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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Verify A (adj A)= (adj A) A = | A | I: \[ \begin{bmatrix} 1&-1&2 \\ 3&0&-2 \\ 1&0&3 \\ \end{bmatrix} \]

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Toolbox:
  • (i)A square matrix is said to be singular if $|A|$ =0.
  • (ii)The adjoint of a 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 of the matrix A is denoted by adj A.
  • Let $A=\begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix}$
  • Then $adj \;A=\begin{bmatrix}A_{11} & A_{12} & A_{13}\\A_{21} & A_{22} & A_{23}\\A_{31} & A_{32} & A_{33}\end{bmatrix}$
Given:For A=$\begin{bmatrix}1 & -1 & 2\\3 & 0 & -2\\1 & 0 & 3\end{bmatrix}$
Verify A(adj A)=(adj A)A=|A|I.
Expanding along $R_1$ we get,
$|A|=1(0\times 3-(-2\times 0))-(-1)(3\times 3-(-2\times 1)+2(3\times 0+0\times 1)$
$\;\;\;\;=0+1(9+2)+0$
$\;\;\;\;=11.$
Hence it is not a singular matrix.
To find the adj A,let us find the minors and cofactors of the respective elements .
$M_{11}=\begin{bmatrix}0 & -2\\0 & 3\end{bmatrix}=0-0=0.$
$M_{12}=\begin{bmatrix}3 & -2\\1 & 3\end{bmatrix}=9+2=11.$
$M_{13}=\begin{bmatrix}3 & 0\\1 & 0\end{bmatrix}=0.$
$M_{21}=\begin{bmatrix}-1 & 2\\0 & 3\end{bmatrix}=-3-0=-3.$
$M_{22}=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix}=3-2=1.$
$M_{23}=\begin{bmatrix}1 & -1\\1 & 0\end{bmatrix}=0-(-1)=1.$
$M_{31}=\begin{bmatrix}-1 & 2\\0 & -2\end{bmatrix}=2-0=2.$
$M_{32}=\begin{bmatrix}1 & 2\\3 & -2\end{bmatrix}=-2-6=-8.$
$M_{33}=\begin{bmatrix}1 & -1\\3 & 0\end{bmatrix}=0--(-3)=3.$
$A_{11}=(-1)^{1+1}.0=0.$
$A_{12}=(-1)^{1+2}.11=-11.$
$A_{13}=(-1)^{1+3}.0=0.$
$A_{21}=(-1)^{2+1}\times -3=3.$
$A_{22}=(-1)^{2+2}.1=1.$
$A_{23}=(-1)^{2+3}.1=-1.$
$A_{31}=(-1)^{3+1}.2=2.$
$A_{32}=(-1)^{3+2}.-8=8.$
$A_{33}=(-1)^{3+3}.3=3.$
Hence $adj \;A=\begin{bmatrix}A_{11} & A_{12} & A_{13}\\A_{21} & A_{22} & A_{23}\\A_{31} & A_{32} & A_{33}\end{bmatrix}$
$\;\;\;\;\qquad\;\;=\begin{bmatrix}0 & 3& 2\\-11 & 1 & 8\\0 & -1 & 3\end{bmatrix}$
Let us find A(adj A).
Matrix multiplication is obtained by multiplying the rows and columns.
(i.e)$\begin{bmatrix}1 & -1& 2\\3 & 0 & -2\\1 & 0 & 3\end{bmatrix}\begin{bmatrix}0 & 3& 2\\-11 & 1 & 8\\0 & -1 & 3\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}0\times 1-1\times -11+2\times 0 & 1\times 3+(-1)\times 1+2\times (-1) & 1\times 2+(-1)\times 8+2\times 3\\0\times 3+0\times -11+(-2)\times 0 & 3\times 3+0\times 1+(-2)\times (-1) & 3\times 2+0\times 8+(-2)\times 3\\1\times 0+0\times -11+3\times 0 & 1\times 3+0\times 1+3\times (-1) & 1\times 2+(0)\times 8+3\times 3\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}0+11+0 & 3-1-2 & 2-8+6\\0+0+0 & 9+0+2 & 6+0-6\\0+0+0& 3+0-1 & 2+0+9\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}11 & 0 & 0\\0 & 11 & 0\\0 & 0 &11\end{bmatrix}$------(1)
Let us find (adj A)A=$\begin{bmatrix}0 & 3 & 2\\-11 & 1 & 8\\0 &-1 & 3\end{bmatrix}\begin{bmatrix}1 & -1 & 2\\3 & 0 & -2\\1 & 0 &3\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}0+3\times 3+2\times 1&0\times -2+3\times 0+2\times 0 & 2\times 0+3\times -2+2\times 3\\-1\times 1+1\times 3+8\times 1&-11\times -1 +1\times 0+(-1)\times 0 &-11\times (-1)+1\times 0+8\times 0\\0\times 1+(-1)\times 3+3\times 1& 0\times -2+-1\times 0+3\times 0 & 0\times 2+-1\times -2+3\times 3\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}0+9+2 & 0+0+0 &0-6+6\\-11+3+8 &11+0+0 &11+0+0\\0-3+3 & 0+0+0& 0+2+9\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}11 & 0 & 11\\0 & 11 & 0\\0 & 0 &11\end{bmatrix}$-------(2)
let us find |A|I
We know $I=\begin{bmatrix}1 & 0 & 0\\0 & 1& 0\\0 & 0&1\end{bmatrix}$
We have already found |A|I=11.
$|A|I=\begin{bmatrix}11 & 0 & 0\\0 & 11& 0\\0 & 0&11\end{bmatrix}$------(3)
From eq(1),eq(2) and eq(3)
We can find A(adj A)=(adj A)A=|A|I.
$\;\;\;=\begin{bmatrix}11 & 0 & 0\\0 & 11& 0\\0 & 0&11\end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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