 Caroline Elliott

2022-01-21

Solve the system
$3{x}_{1}+2{x}_{2}-{x}_{3}=4$
${x}_{1}-2{x}_{2}+2{x}_{3}=1$
$11{x}_{1}+2{x}_{2}+{x}_{3}=14$ Damian Roberts

Expert

Step 1
Given:
$3{x}_{1}+2{x}_{2}-{x}_{3}=4$
${x}_{1}-2{x}_{2}+2{x}_{3}=1$
$11{x}_{1}+2{x}_{2}+{x}_{3}=14$
Step 2
$\left[\begin{array}{ccc}3& 2& -1\\ 1& -2& 2\\ 11& 2& 1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}4\\ 1\\ 14\end{array}\right]$
Therefore augmented matrix is
$\left[\begin{array}{cccc}3& 2& -1& 4\\ 1& -2& 2& 1\\ 11& 2& 1& 14\end{array}\right]$
Step 3
Row transformations

$\left[\begin{array}{cccc}3& 2& -1& 4\\ 0& -8& 7& -1\\ 0& -16& 14& -2\end{array}\right]$
${R}_{3}\to {R}_{3}-2{R}_{2}$
$\left[\begin{array}{cccc}3& 2& -1& 4\\ 0& -8& 7& 7\\ 0& 0& 0& 0\end{array}\right]$
Step 4
Thus,
${x}_{3}$ is free variable
${x}_{2}=\frac{-1}{8}\left(-1-7{x}_{3}\right)$
${x}_{1}=\frac{1}{4}\left(5-{x}_{3}\right)$
for ${x}_{3}=1$
${x}_{2}=1$
${x}_{1}=1$
Therefore, $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ is one of the solution of given system. enguinhispi

Expert

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

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