Question

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

Answer 1

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

$A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$
$\therefore |A|=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]=2(-4+4)-(-3)(6+4)+5(3-2)$
$=0-6+5=-1\ne 0$
Thus, A is non-singular. Therefore, its inverse exists.
Therefore, the given system is consistent and has a unique solution given by $X=A^{-1}B$.
Cofactors of A are
$A_{11}=-4+4=0,A_{12}=-(-6+4)=2,A_{13}=3-2=1A_{21}=-(6-5)=-1,A_{22}-4-5=-9,A_{23}=-(2+3)=-5A_{31}=(12-0)=2,A_{32}=-(-8-15)=23,A_{33}=4+9=13$
$adj\left(A\right)={\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]}^T=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}\left(adj\right)\left(A\right)=\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$
Now, $X=A^{-1}B\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]-\left[\begin{array}{c}0-5+6\\ -22-45+69\\ -11-25+39\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
Hence, x=1, y=2 and z=3