Question

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

Answer 1

$|A|=detA=\left|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right|=2(-4+4)+3(-6+4)+5(3-2)=-6+5=-1$

Now, finding the cofactors of the matrix $A$:

$A_{11}=\left|\begin{array}{cc}2& -4\\ 1& -2\end{array}\right|=0$

$A_{12}=-\left|\begin{array}{cc}3& -4\\ 1& -2\end{array}\right|=2$

$A_{13}=\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|=1$

$A_{21}=-\left|\begin{array}{cc}-3& 5\\ 1& -2\end{array}\right|=-1$

$A_{22}=\left|\begin{array}{cc}2& 5\\ 1& -2\end{array}\right|=-9$

$A_{23}=-\left|\begin{array}{cc}2& -3\\ 1& 1\end{array}\right|=-5$

$A_{31}=\left|\begin{array}{cc}-3& 5\\ 2& -4\end{array}\right|=2$

$A_{32}=-\left|\begin{array}{cc}2& 5\\ 3& -4\end{array}\right|=23$

$A_{33}=\left|\begin{array}{cc}2& -3\\ 3& 2\end{array}\right|=13$

Adjoint of matrix $A$ will be given by:

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

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

The system of equations can be reduced in the matrix form $AX=B$ as

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

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

Therefore $x=1,y=2,z=3$ is the solution to the system of equations.