Question

If $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$, find $A^{-1}$. Use it to solve the system of equations.
$2x-3y+5z=11$
$3x+2y-4z=-5$
$x+y-2z=-3$.

Answer 1

$Co\left(A\right)=\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]$

We know that $Adj\left(A\right)=TransposeofCo\left(A\right)$

$\therefore Adj\left(A\right)=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]$

Since, $A^{-1}=\frac{1}{|A|}Adj\left(A\right)$ and $|A|=-1$

$\therefore A^{-1}=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$

Now, $Ax=b\Rightarrow x=A^{-1}\times b$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$

Therefore, $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

Hence, $x=1,y=2$ and $z=3$.