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

$A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right],\left|A\right|=2(-4+4)+3(-6+4)+5(3-2)=-6+5=-1AdjA=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\therefore A^{-1}=\frac{AdjA}{\left|A\right|}=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]2x-3y+5z=113x+2y-4z=-5x+y-2z=-3$

$\therefore$ Let $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]\therefore AX=B\Rightarrow X=A^{-1}B\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\therefore x=1,y=2,z=3$