Question

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

Answer 1

Here, $|A|=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1\ne 0\therefore A^{-1}\text{exists}Now,A_{11}={(-1)}^{1+1}\left|\begin{array}{cc}2& -4\\ 1& -2\end{array}\right|=+(-4+4)=0A_{12}={(-1)}^{1+2}\left|\begin{array}{cc}3& -4\\ 1& -2\end{array}\right|=-(-6+4)=2A_{13}={(-1)}^{1+3}\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|=+(3-2)=1A_{21}={(-1)}^{2+1}\left|\begin{array}{cc}-3& 5\\ 1& -2\end{array}\right|=-(6-5)=-1A_{22}={(-1)}^{2+2}\left|\begin{array}{cc}2& 5\\ 1& -2\end{array}\right|=+(-4-5)=-9A_{23}={(-1)}^{2+3}\left|\begin{array}{cc}2& -3\\ 1& 1\end{array}\right|=-(2+3)=-5A_{31}={(-1)}^{3+1}\left|\begin{array}{cc}-3& 5\\ 2& -4\end{array}\right|=+(12-10)=2A_{32}={(-1)}^{3+2}\left|\begin{array}{cc}2& 5\\ 3& -4\end{array}\right|=-(-8-15)=23A_{33}={(-1)}^{3+3}\left|\begin{array}{cc}2& -3\\ 3& 2\end{array}\right|=+(4+9)=13\therefore A^{-1}=\frac{1}{|A|}\left(AdjA\right)=\frac{1}{-1}{\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]}^T=\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$
The given system of equations can be written as:$\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]i.e.,AX=B$
$\therefore$ Solution of the given system is given by $X=A^{-1}B.\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]=\left[\begin{array}{c}0-5+6\\ -22-45+69\\ -11-25+39\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$ Hence, $x=1,y=2$ and $z=3$