Solve each of the following systems of equations using matrix method. $x + 2y + z = 7$; $x + 3z = 11$; $2x

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Solve each of the following systems of equations using matrix method.
$x + 2y + z = 7$ ;
$x + 3z = 11$ ;
$2x - 3y = 1$ .

Solve each of the following systems of equations using matrix method.
$x + 2y + z = 7$ ;
$x + 3z = 11$ ;
$2x - 3y = 1$ .

Solution:

We need to find: $-\mathrm{x}, \mathrm{y}, \mathrm{z}$
The given set of lines are : –
$\begin{array}{l} x+2 y+z=7 \\ x+3 z=11 \\ 2 x-3 y=1 \end{array}$
Now converting following equations in matrix form,
$\mathrm{AX}=\mathrm{B}$
$\left(\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 7 \\ -11 \\ 1 \end{array}\right)$
$\mathrm{R}_{2}-\mathrm{R}_{1}$
$\begin{array}{l} \mathrm{R}_{3}-2 \mathrm{R}_{1} \\ \left(\begin{array}{ccc} 1 & 2 & 1 \\ 0 & -2 & 2 \\ 0 & -7 & -2 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 7 \\ 4 \\ -13 \end{array}\right) \\ \mathrm{R}_{3}+\mathrm{R}_{2} \\ \left(\begin{array}{ccc} 1 & 2 & 1 \\ 0 & -2 & 2 \\ 0 & -9 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 7 \\ 4 \\ -9 \end{array}\right) \end{array}$
Again on converting back into equations we obtain
$\begin{array}{l} x+2 y+z=7 \\ -2 y+2 z=4 \\ -9 y=-9 \\ y=1 \\ -2 \times 1+2 z=4 \\ 2 z=6 \\ z=3 \end{array}$
$\begin{array}{l} x+2 \times 1+3=7 \\ x=7-2-3 \\ x=2 \\ \therefore x=2, y=1, z=3 \end{array}$