Question

If $A=$ $\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$, then find $A^{-1}$.
Use it to solve the system of equations
$2x-3y+5z=113x+2y-4z=-5x+y-2z=-3$

Answer 1

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

$|A|=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1\ne 0$

$\therefore A$ is non-singular so its inverse exists.
$C_{ij}$ are the cofactors of the determinant,

$C_{11}=(-1{)}^2\left[\begin{array}{cc}2& -4\\ 1& -2\end{array}\right]=0$

$C_{12}=(-1{)}^3\left[\begin{array}{cc}3& -4\\ 1& -2\end{array}\right]=2$

$C_{13}=(-1{)}^4\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]=1$

$C_{21}=(-1{)}^3\left[\begin{array}{cc}-3& 5\\ 1& -2\end{array}\right]=-1$

$C_{22}=(-1{)}^4\left[\begin{array}{cc}2& 5\\ 1& -2\end{array}\right]=-9$

$C_{23}=(-1{)}^5\left[\begin{array}{cc}2& -3\\ 1& 1\end{array}\right]=-5$

$C_{31}=(-1{)}^4\left[\begin{array}{cc}-3& 5\\ 2& -4\end{array}\right]=2$

$C_{32}=(-1{)}^5\left[\begin{array}{cc}2& 5\\ 3& -4\end{array}\right]=23$

$C_{33}=(-1{)}^6\left[\begin{array}{cc}2& -3\\ 3& 2\end{array}\right]=13$

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

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

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

Also, $2x-3y+5z=113x+2y-4z=-5x+y-2z=-3$

The given system of equations can be written as $AX=B$, where

$A$= $\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\text{and}B=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$

So, $X$ = $A^{-1}B$

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

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0-5+6\\ -22-45+69\\ -11-25+39\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

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