Solve the system 3x1+2x2−x3=4 x1−2x2+2x3=1 11x1+2x2+x3=14

Step 1
Consider the provided question,
The given system of equation is,
${\displaystyle x_1-2x_2+2x_3=1}$ (I)
${\displaystyle 3x_1+2x_2-x_3=4}$ (II)
${\displaystyle 11x_1+2x_2+x_3=14}$ (III)
Operate ${\displaystyle \left(II\right)-(\left(I\right)\times 3)}$ and ${\displaystyle \left(III\right)-(\left(I\right)\times 11)}$ we obtain,
${\displaystyle 8x_2-7x_3=1}$ (IV)
And, ${\displaystyle 24x_2-21x_3=3\Rightarrow 8x_2-7x_3=1}$ (V)
From (IV) and (V),
${\displaystyle 0x_2-0x_3=0}$
Thus, the system of equation is written as,
${\displaystyle x_1-2x_2+2x_3=1}$
${\displaystyle 8x_2-7x_3=1}$
${\displaystyle 0x_2-0x_3=0}$
And, the corresponding coefficient matrix is,
${\displaystyle \left[\begin{array}{ccc}1& -1& 2\\ 0& 8& -7\\ 0& 0& 0\end{array}\right]}$
Step 2
Since, the corresponding coefficient matrix in row echelon form,
And, the required equivalent system is,
${\displaystyle x_1-2x_2+2x_3=1}$
${\displaystyle 8x_2-7x_3=1}$
${\displaystyle 0x_2-0x_3=0}$
Augumented matrix:
${\displaystyle \left[\begin{array}{cccc}1& -2& 2& 1\\ 0& 8& -7& 1\\ 0& 0& 0& 0\end{array}\right]}$
Since
${\displaystyle \text{Rank}=\text{number of non}=\text{zero row in row echelon form of matrix}}$
So, rank of coefficient matrix is 2 and, that of augumented matrix is also 2. NKS Therefore, the given system of equation is consistent.
Since, number of free variable
${\displaystyle =\text{number of unknown variable}-\text{rank of coefficient matrix}}$
${\displaystyle =3-2}$
${\displaystyle =1}$
So, the system has one free variable.
let it be ${\displaystyle x_3}$ and ${\displaystyle x_3=a;a\in R}$
Step 3
${\displaystyle x_1-2x_2+2a=1}$
${\displaystyle 8x_2-7a=1}$
${\displaystyle 0x_2-0a=0}$
${\displaystyle \Rightarrow x_2=\frac{1+7a}{8}}$
Now,