# Q : 4          Verify $\small A (adj A)=(adjA)A=|A| I$.                  $\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Given matrix: $\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Let  $\small A= \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1} \begin{vmatrix} 0 &-2 \\ 0& 3 \end{vmatrix} = 0$

$\small A_{12} = (-1)^{1+2} \begin{vmatrix} 3 &-2 \\1& 3 \end{vmatrix} = -(9+2) =-11$

$\small A_{13} = (-1)^{1+3} \begin{vmatrix} 3 &0 \\ 1& 0 \end{vmatrix} = 0$

$\small A_{21} = (-1)^{2+1} \begin{vmatrix} -1 &2 \\ 0& 3 \end{vmatrix} = -(-3-0)= 3$

$\small A_{22} = (-1)^{2+2} \begin{vmatrix} 1 &2 \\ 1& 3 \end{vmatrix} = 3-2=1$

$\small A_{23} = (-1)^{2+3} \begin{vmatrix} 1 &-1 \\ 1& 0 \end{vmatrix} = -(0+1) = -1$

$\small A_{31} = (-1)^{3+1} \begin{vmatrix} -1 &2 \\ 0& -2 \end{vmatrix} = 2$

$\small A_{32} = (-1)^{3+2} \begin{vmatrix} 1 &2 \\ 3& -2 \end{vmatrix} = -(-2-6) = 8$

$\small A_{33} = (-1)^{3+3} \begin{vmatrix} 1 &-1 \\ 3& 0 \end{vmatrix} = 0+3 =3$

Hence, $\small adjA = \begin{bmatrix} 0 &3 &2 \\ -11 & 1& 8\\ 0 &-1 & 3 \end{bmatrix}$

Now,

$\small A (adj A) =\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+11+0 &3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 &3+0-3 & 2+0+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

also,

$\small A (adj A) =\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+9+2 &0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 &0+0+0 & 0+2+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = 1(0-0) +1(9+2) +2(0-0) = 11$

So, $\small |A|I = 11\begin{bmatrix} 1 &0&0 \\ 0& 1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11 &0&0 \\ 0& 11&0\\ 0&0&11 \end{bmatrix}$

Hence we get,

$\small A (adj A)=(adj A)A=|A|I$.

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