# Q : 8        Let $\dpi{100} A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that,                (i) $\dpi{100} [adj A]^-^1 = adj (A^-^1)$

D Divya Prakash Singh

Given that $\dpi{100} A=\begin{bmatrix} 1 &2 &1 \\ 2 &3 &1 \\ 1 & 1 & 5 \end{bmatrix}$;

So, let us assume that $A^{-1} = B$ matrix and $adjA = C$ then;

$|A| = 1(15-1) -2(10-1) +1(2-3) = 14-18-1 = -5 \neq 0$

Hence its inverse exists;

$A^{-1} = \frac{1}{|A|} adjA$    or  $B = \frac{1}{|A|}C$;

so, we now calculate the value of $adjA$

Cofactors of A;

$A_{11}= (-1)^{1+1}(15-1) = 14$          $A_{12}= (-1)^{1+2}(10-1) = -9$

$A_{13}= (-1)^{1+3}(2-3) = -1$          $A_{21}= (-1)^{2+1}(10-1) = -9$

$A_{22}= (-1)^{2+2}(5-1) = 4$             $A_{23}= (-1)^{2+3}(1-2) = 1$

$A_{31}= (-1)^{3+1}(2-3) = -1$         $A_{32}= (-1)^{3+2}(1-2) = 1$???????

$A_{33}= (-1)^{3+3}(3-4) = -1$???????

$\Rightarrow adjA =C= \begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

$A^{-1} =B= \frac{1}{|A|} adjA = \frac{1}{-5}\begin{bmatrix} 14 &-9 &-1 \\ -9& 4& 1\\ -1& 1 &-1 \end{bmatrix}$

Finding the inverse of C;

$|C| = 14(-4-1)+9(9+1)-1(-9+4) = -70+90+5 = 25 \neq 0$

Hence its inverse exists;

$C^{-1} = \frac{1}{|C|}adj C$

Now, finding the $adjC$;

$C_{11}= (-1)^{1+1}(-4-1) = -5$                       $C_{12}= (-1)^{1+2}(9+1) = -10$

$C_{13}= (-1)^{1+3}(-9+4) = -5$                       $C_{21}= (-1)^{2+1}(9+1) = -10$

$C_{22}= (-1)^{2+2}(-14-1) = -15$                 $C_{23}= (-1)^{2+3}(14-9) = -5$

$C_{31}= (-1)^{3+1}(-9+4) = -5$                        $C_{32}= (-1)^{3+2}(14-9) = -5$

$C_{33}= (-1)^{3+3}(56-81) = -25$

$adjC = \begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix}$

$C^{-1} = \frac{1}{|C|}adjC = \frac{1}{25}\begin{bmatrix} -5 &-10 &-5 \\ -10& -15 & -5\\ -5& -5& -25 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

or $L.H.S. = C^{-1} = [adjA]^{-1} = \begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Now, finding the R.H.S.

$adj (A^{-1}) = adj B$

$A^{-1} =B= \begin{bmatrix} \frac{-14}{5} &&\frac{9}{5} &&\frac{1}{5} \\ \\ \frac{9}{5}&& \frac{-4}{5}&& \frac{-1}{5}\\ \\ \frac{1}{5}&& \frac{-1}{5} &&\frac{1}{5}\end{bmatrix}$

Cofactors of B;

$B_{11}= (-1)^{1+1}(\frac{-4}{25}-\frac{1}{25}) = \frac{-1}{5}$

$B_{12}= (-1)^{1+2}(\frac{9}{25}+\frac{1}{25}) =- \frac{2}{5}$

$B_{13}= (-1)^{1+3}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{21}= (-1)^{2+1}(\frac{9}{25}+\frac{1}{25}) = -\frac{2}{5}$

$B_{22}= (-1)^{2+2}(\frac{-14}{25}-\frac{1}{25}) = \frac{-3}{5}$

$B_{23}= (-1)^{2+3}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{31}= (-1)^{3+1}(\frac{-9}{25}+\frac{4}{25}) = \frac{-1}{5}$

$B_{32}= (-1)^{3+2}(\frac{14}{25}-\frac{9}{25}) = \frac{-1}{5}$

$B_{33}= (-1)^{3+3}(\frac{56}{25}-\frac{81}{25}) = -1$

$R.H.S. = adjB = adj(A^{-1}) =\begin{bmatrix} -\frac{1}{5} && -\frac{2}{5} &&-\frac{1}{5} \\ \\ -\frac{2}{5}&& -\frac{3}{5} && -\frac{1}{5}\\ \\ -\frac{1}{5} && -\frac{1}{5} && -1 \end{bmatrix}$

Hence L.H.S. = R.H.S. proved.

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