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

Given system of equations

$2x-3y+5z=11$
$3x+2y-4z=5$

$x+y-2z=3$

This can be written as

$AX=B$

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

Here, $|A|=2(-4+4)+3(-6+4)+5(3-2)$

$\Rightarrow |A|=-6+5=-1$

Since, $|A|\ne 0$

Hence, the system of equations is consistent and has a unique solution given by $X==A^{-1}B$

$A^{-1}=\frac{adjA}{|A|}$ and $adjA=C^T$

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

$\Rightarrow C_{11}=-4+4=0$

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

$\Rightarrow C_{12}=-(-6+4)=2$

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

$\Rightarrow C_{13}=3-2=1$

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

$\Rightarrow C_{21}=-(6-5)=-1$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}2& 5\\ 1& -2\end{array}\right|$

$\Rightarrow C_{22}=-4-5=-9$

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

$\Rightarrow C_{23}=-(2+3)=-5$

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

$\Rightarrow C_{31}=12-10=2$

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

$\Rightarrow C_{32}=-(-8-15)=23$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}2& -3\\ 3& 2\end{array}\right|$

$\Rightarrow C_{33}=4+9=13$

Hence, the co-factor matrix is $C=\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]$

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

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

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

Solution is given by

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

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

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}8\\ 4\\ 12\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ -46\\ -25\end{array}\right]$

Hence, $x=-1,y=-46,z=-25$