Solve each of the following systems of equations using matrix method. $x + y + z = 6$; $x + 2z = 7$; $3x + y + z = 12$.

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Solve each of the following systems of equations using matrix method.
$x + y + z = 6$ ;
$x + 2z = 7$ ;
$3x + y + z = 12$ .

Solve each of the following systems of equations using matrix method.
$x + y + z = 6$ ;
$x + 2z = 7$ ;
$3x + y + z = 12$ .

Solution:

We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$
The given set of lines are : –
$x+y+z=6$
$\begin{array}{l} x+2 z=7 \\ 3 x+y+z=12 \end{array}$
Now converting following equations in matrix form,
$\begin{array}{l} \mathrm{AX}=\mathrm{B} \\ \left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 6 \\ 7 \\ 12 \end{array}\right) \end{array}$
$\begin{array}{l} \mathrm{R}_{2}-\mathrm{R}_{1} \\ \mathrm{R}_{3}-3 \mathrm{R}_{1} \end{array}$
$\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & -2 & -2 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 6 \\ 1 \\ -6 \end{array}\right)$
$\mathrm{R}_{3}+2 \mathrm{R}_{2}$
$\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & -4 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 6 \\ 1 \\ -4 \end{array}\right)$
Again on converting back into equations we obtain
$\begin{array}{l} x+y+z=6 \\ -y+z=1 \\ -4 y=-4 \\ y=1 \\ y+1+z=1 \\ z=2 \\ x+1+2=6 \\ x=6-3 \end{array}$
$x=3$
$\therefore x=3, y=1, z=2$