Question

Find the inverse of the matrix $\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& 5\\ 2& 4& 7\end{array}\right]$ by adjoint method.

Answer 1

$\left[\begin{array}{ccc}1& 2& 3\\ 1& 1& 5\\ 2& 4& 7\end{array}\right]$

The co-factor matrix

$C_{11}={(-1)}^{1+1}{M}_{11}=\left|\begin{array}{cc}1& 5\\ 4& 7\end{array}\right|=7-20=-13$

$C_{12}={(-1)}^{1+2}{M}_{12}=-\left|\begin{array}{cc}1& 5\\ 2& 7\end{array}\right|=-(7-10)=3$

$C_{13}={(-1)}^{1+3}{M}_{13}=\left|\begin{array}{cc}1& 1\\ 2& 4\end{array}\right|=4-2=2$

$C_{21}={(-1)}^{2+1}{M}_{21}=-\left|\begin{array}{cc}2& 3\\ 4& 7\end{array}\right|=-(14-12)=-2$

$C_{22}={(-1)}^{2+2}{M}_{22}=\left|\begin{array}{cc}1& 3\\ 2& 7\end{array}\right|=7-6=1$

$C_{23}={(-1)}^{2+3}{M}_{23}=-\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=-(4-4)=0$

$C_{31}={(-1)}^{3+1}{M}_{31}=\left|\begin{array}{cc}2& 3\\ 1& 5\end{array}\right|=10-3=7$

$C_{32}={(-1)}^{3+2}{M}_{32}=-\left|\begin{array}{cc}1& 3\\ 1& 5\end{array}\right|=-(5-3)=-2$

$C_{33}={(-1)}^{3+3}{M}_{33}=\left|\begin{array}{cc}1& 2\\ 1& 1\end{array}\right|=1-2=-1$

Since the transpose of the co-factor matrix of $A$ is adj$A$

$\Rightarrow$ Adj$\left(A\right)$

$={\left|\begin{array}{ccc}-13& 3& 2\\ -2& 1& 0\\ 7& -2& -1\end{array}\right|}^T$

$=\left|\begin{array}{ccc}-13& -2& 7\\ 3& 1& -2\\ 2& 0& -1\end{array}\right|$

Since det$A$ from the above matrix $A$

$=1(7-20)-2(7-10)+3(4-2)=-13+6+6=-1\ne 0$

Hence inverse of $A^{-1}$ exists.

$A^{-1}=\frac{adjA}{\left|A\right|}$

$=-\left|\begin{array}{ccc}-13& -2& 7\\ 3& 1& -2\\ 2& 0& -1\end{array}\right|$

$=\left|\begin{array}{ccc}13& 2& -7\\ -3& -1& 2\\ -2& 0& 1\end{array}\right|$