Question

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

Answer 1

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

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

$A^{-1}=\frac{\left[\begin{array}{ccc}(2\times 2-3\times 1)& -(2\times 2-1\times 1)& (2\times 3-2\times 1)\\ -(3\times 2-3\times 1)& (1\times 2-1\times 1)& -(1\times 3-3\times 1)\\ (3\times 3-2\times 1)& -(1\times 1-2\times 1)& (1\times 2-2\times 3)\end{array}\right]}{|1(2\times 2-3\times 1)-3(2\times 2-1\times 1)+1(2\times 3-2\times 1)|}$

$A^{-1}=\frac{\left[\begin{array}{ccc}1& -3& 4\\ -3& 1& 0\\ 1& 1& -4\end{array}\right]}{1-9+4}$

$A^{-1}=\frac{\left[\begin{array}{ccc}1& -3& 4\\ -3& 1& 0\\ 1& 1& -4\end{array}\right]}{-4}$

$A^{-1}=\left[\begin{array}{ccc}\frac{-1}{4}& \frac{3}{4}& -1\\ \frac{3}{4}& \frac{-1}{4}& 0\\ \frac{-1}{4}& \frac{-1}{4}& 1\end{array}\right]$