Question

If $A=\left[\begin{array}{ccc}1& -1& 0\\ 2& 5& 3\\ 0& 2& 1\end{array}\right]$, find $A^{-1}$

Answer 1

$A=\left[\begin{array}{ccc}1& -1& 0\\ 2& 5& 3\\ 0& 2& 1\end{array}\right]$

$\left|A\right|=1(5-6)+1(2-0)+0=-1+2=1\ne 0$

$\therefore A^{-1}$ exists.

$M_{11}=\left|\begin{array}{cc}5& 3\\ 2& 1\end{array}\right|=5-6=-1\Rightarrow C_{11}={(-1)}^{1+1}{M}_{11}={(-1)}^2\times -1=-1$

$M_{12}=\left|\begin{array}{cc}2& 3\\ 0& 1\end{array}\right|=2-0=2\Rightarrow C_{12}={(-1)}^{1+2}\times 2={(-1)}^3\times 2=-2$

$M_{13}=\left|\begin{array}{cc}2& 5\\ 0& 2\end{array}\right|=4-0=4\Rightarrow C_{13}={(-1)}^{1+3}={(-1)}^4\times 4=4$

$M_{21}=\left|\begin{array}{cc}-1& 0\\ 2& 1\end{array}\right|=-1-0=-1\Rightarrow C_{21}={(-1)}^{2+1}\times -1={(-1)}^3\times -1=1$

$M_{22}=\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1-0=1\Rightarrow C_{22}={(-1)}^{2+2}\times -1={(-1)}^3\times -1=1$

$M_{23}=\left|\begin{array}{cc}1& -1\\ 0& 2\end{array}\right|=2-0=2\Rightarrow C_{23}={(-1)}^{2+3}\times -1={(-1)}^5\times -1=1$

$M_{31}=\left|\begin{array}{cc}-1& 0\\ 5& 3\end{array}\right|=-3-0=-3\Rightarrow C_{31}={(-1)}^{3+1}\times -3={(-1)}^4\times -1=1$

$M_{32}=\left|\begin{array}{cc}1& 0\\ 2& 3\end{array}\right|=3-0=3\Rightarrow C_{32}={(-1)}^{3+2}\times 3={(-1)}^5\times 3=-3$

$M_{33}=\left|\begin{array}{cc}1& -1\\ 2& 5\end{array}\right|=5+2=7\Rightarrow C_{33}={(-1)}^{3+3}\times 7={(-1)}^6\times 7=7$

Cofactor matrix$=\left[\begin{array}{ccc}-1& -2& 4\\ 1& 1& 1\\ 1& -3& 7\end{array}\right]$

$Adj\left\{A\right\}={\left[C_{ij}\right]}^T={\left[\begin{array}{ccc}-1& -2& 4\\ 1& 1& 1\\ 1& -3& 7\end{array}\right]}^T$

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

$A^{-1}=\frac{Adj\left\{A\right\}}{\left|A\right|}=\frac{1}{1}\left[\begin{array}{ccc}-1& 1& 1\\ -2& 1& -3\\ 4& 1& 7\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{ccc}-1& 1& 1\\ -2& 1& -3\\ 4& 1& 7\end{array}\right]$