If $A =
\begin{bmatrix}
3 & 1 & 2 \\
3 & 2 & -3 \\
2 & 0 & -1
\end{bmatrix}$, find $A^{-1}$.
Hence solve the system of equations
$3x+3y+2z=1$, $x+2y=4$, $2x-3y-z=5$
$\left|A\right| =$
$
\begin{vmatrix}
3 & 1 & 2 \\
3 & 2 & -3 \\
2 & 0 & -1
\end{vmatrix} = 3 \times \left(-2\right) -1 \times \left(-3+6\right) + 2 \times \left(-4\right) = -17 \neq 0
$
Therefore, $A^{-1}$ exists.
Cofactors of A are:
$A_{11} = \left(-1\right)^{1+1}$
$\begin{vmatrix}
2 & -3 \\
0 & -1
\end{vmatrix}=-2$
$A_{12} = \left(-1\right)^{1+2}$
$\begin{vmatrix}
3 & -3 \\
2 & -1
\end{vmatrix}=- \left(-3+6\right)=-3$
$A_{13} = \left(-1\right)^{1+3}$
$\begin{vmatrix}
3 & 2 \\
2 & 0
\end{vmatrix}=-4$
$A_{21} = \left(-1\right)^{2+1}$
$\begin{vmatrix}
1 & 2 \\
0 & -1
\end{vmatrix}=-\left(-1\right)=1$
$A_{22} = \left(-1\right)^{2+2}$
$\begin{vmatrix}
3 & 2 \\
2 & -1
\end{vmatrix}=-3-4=-7$
$A_{23} = \left(-1\right)^{2+3}$
$\begin{vmatrix}
3 & 1 \\
2 & 0
\end{vmatrix}=- \left(-2\right)=2$
$A_{31} = \left(-1\right)^{3+1}$
$\begin{vmatrix}
1 & 2 \\
2 & -3
\end{vmatrix}=-3-4=-7$
$A_{32} = \left(-1\right)^{3+2}$
$\begin{vmatrix}
3 & 2 \\
3 & -3
\end{vmatrix}=-\left(-9-6\right)=15$
$A_{33} = \left(-1\right)^{3+3}$
$\begin{vmatrix}
3 & 1 \\
3 & 2
\end{vmatrix}=6-3=3$
$\text{adj A}=$
$\begin{bmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{bmatrix}^t = $
$\begin{bmatrix}
-2 & -3 & -4 \\
1 & -7 & 2 \\
-7 & 15 & 3
\end{bmatrix}^t=$
$
\begin{bmatrix}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{bmatrix}
$
$A^{-1} = \dfrac{\text{adj A}}{\left|A\right|} = \dfrac{-1}{17}$
$
\begin{bmatrix}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{bmatrix}
$
Let $A_1 =$
$\begin{bmatrix}
3 & 3 & 2 \\
1 & 2 & 0 \\
2 & -3 & -1
\end{bmatrix}$,
$B=$
$\begin{bmatrix}
1 \\
4 \\
5
\end{bmatrix}$ and $X=$
$\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$
Then $A_1 X = B \implies X = A_1^{-1}B$
But $A_1 = A^t$ (Comparing A and $A_1$)
Therefore, $X = \left(A^t\right)^{-1}B = \left(A^{-1}\right)^t B$ $\left[\text{Note: } \left(A^t\right)^{-1} = \left(A^{-1}\right)^t\right]$
$X= \dfrac{-1}{17}$
$\begin{bmatrix}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{bmatrix}^t$
$\begin{bmatrix}
1 \\
4 \\
5
\end{bmatrix} = \dfrac{-1}{17}$
$\begin{bmatrix}
-2 & -3 & -4 \\
1 & -7 & 2 \\
-7 & 15 & 3
\end{bmatrix}$
$\begin{bmatrix}
1 \\
4 \\
5
\end{bmatrix}$
$= \dfrac{-1}{17}$
$\begin{bmatrix}
-2-12-20 \\
1-28+10 \\
-7+60+15
\end{bmatrix}$
$= \dfrac{-1}{17}$
$\begin{bmatrix}
-34 \\
-17 \\
68
\end{bmatrix}$
i.e.
$\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} = $
$\begin{bmatrix}
2 \\
1 \\
-4
\end{bmatrix}$
$\implies x=2 \,, y =1 \,, z=-4$