Question

If $A=\left[\begin{array}{ccc}3& 1& 2\\ 3& 2& -3\\ 2& 0& -1\end{array}\right]$, find $A^{-1}$
Hence, solve the system of equation:
$3x+3y+2z=1$
$x+2y=4$
$2x-3y-z=5$

Answer 1

$A=\left[\begin{array}{ccc}3& 1& 2\\ 3& 2& -3\\ 2& 0& -1\end{array}\right]$

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

$|A|=3[-2+0]-1\{-3+6\}+2\{0-4\}$

$=-6-3-8$

$=-17$

$adjA=\left[\begin{array}{ccc}+\left|\begin{array}{cc}2& -3\\ 0& -1\end{array}\right|& -\left|\begin{array}{cc}3& -3\\ 2& -1\end{array}\right|& +\left|\begin{array}{cc}3& 2\\ 2& 0\end{array}\right|\\ -\left|\begin{array}{cc}1& 2\\ 0& -1\end{array}\right|& +\left|\begin{array}{cc}3& 2\\ 2& -1\end{array}\right|& -\left|\begin{array}{cc}3& 1\\ 2& 0\end{array}\right|\\ +\left|\begin{array}{cc}1& 2\\ 2& -3\end{array}\right|& -\left|\begin{array}{cc}3& 2\\ 3& -3\end{array}\right|& +\left|\begin{array}{cc}3& 1\\ 3& 2\end{array}\right|\end{array}\right]$

$=\left[\begin{array}{ccc}-2& -3& -4\\ 1& -7& 2\\ -7& 15& 3\end{array}\right]$

$=\left[\begin{array}{ccc}-2& 1& -7\\ -3& -7& 15\\ -4& 2& 3\end{array}\right]$

$A^{-1}=\frac{1}{-17}\left[\begin{array}{ccc}-2& 1& -7\\ -3& -7& 15\\ -4& 2& 3\end{array}\right]$

$3x+3y+2z=1-----\left(1\right)$

$x+2y=4\to x=4-2y$

$2x-3y-z=5---\left(3\right)$

Sub $x$ in $\left(3\right)$

$2(4-2y)-3y-z=5$

$8-4y-3y-z=5$

$8-7y-z=5$

$8-7y-5=z$

$z=3-7y$

Sub $x$ and $z$ in $\left(1\right)$

$3(4-2y)+3y+2(9-7y)=1$

$12-6y+3y+6-14y=1$

$-17y+17=0$

$-17y=-17$

$y=1$

$x=4-2y$

$x=4-2\left(1\right)$

$x=2$

$z=3-7y$

$z=3-7\left(1\right)$

$z=-4$