Question
Solve system of linear equation, using matrix method :
$\begin{aligned}
x+y+z & =6 \\
y+3 z & =11 \\
x+z & =2 y
\end{aligned}$

Answer

Let first, second and third numbers be denoted by $x, y$ and $z$, respectively.
Then, according to given conditions, we have
$\begin{array}{l}
x+y+z=6 \\
y+3 z=11 \\
x+z=2 y \text { or } x-2 y+z=0
\end{array}$
This system can be written as $AX = B$, where,
$A =\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$
Hence, $| A |=1(1+6)-1(0-3)+1(0-1)=9 \neq 0$.
Now we find adj A.
$\begin{array}{l}
A_{11}=1(1+6)=7, \\
A_{12}=-(0-3)=3, \\
A_{13}=-1 \\
A_{21}=-(1+2)=-3,
\end{array}$
$\begin{array}{l}A_{22}=0 \\ A_{23}=-(-2-1)=3 \\ A_{31}=(3-1)=2, A_{32}=-(3-0)=-3, \\ A_{33}=(1-0)=1\end{array}$
Hence, adj $A =\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
Thus, $\quad A ^{-1}=\frac{1}{|A|} \operatorname{adj}( A )$
$=\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
Since $\quad X=A^{-1} B$
$X =\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$
or $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}42-33+0 \\ 18+0+0 \\ -6+33+0\end{array}\right]$
$=\frac{1}{9}\left[\begin{array}{c}9 \\ 18 \\ 27\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
Thus, $\quad x=1, y=2, z=3$.

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