Caroline Elliott

Answered

2022-01-21

Solve the system

$3{x}_{1}+2{x}_{2}-{x}_{3}=4$

${x}_{1}-2{x}_{2}+2{x}_{3}=1$

$11{x}_{1}+2{x}_{2}+{x}_{3}=14$

Answer & Explanation

Damian Roberts

Expert

2022-01-22Added 14 answers

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

$R}_{2}\to 3{R}_{2}-{R}_{1},\text{}{R}_{3}\to 3{R}_{3}-11{R}_{1$

$\left[\begin{array}{cccc}3& 2& -1& 4\\ 0& -8& 7& -1\\ 0& -16& 14& -2\end{array}\right]$

$R}_{3}\to {R}_{3}-2{R}_{2$

$\left[\begin{array}{cccc}3& 2& -1& 4\\ 0& -8& 7& 7\\ 0& 0& 0& 0\end{array}\right]$

Step 4

Thus,

$x}_{3$ is free variable

${x}_{2}=\frac{-1}{8}(-1-7{x}_{3})$

${x}_{1}=\frac{1}{4}(5-{x}_{3})$

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.

Given:

Step 2

Therefore augmented matrix is

Step 3

Row transformations

Step 4

Thus,

for

Therefore,

enguinhispi

Expert

2022-01-23Added 15 answers

Step 1

Consider the provided question,

The given system of equation is,

Operate

And,

From (IV) and (V),

Thus, the system of equation is written as,

And, the corresponding coefficient matrix is,

Step 2

Since, the corresponding coefficient matrix in row echelon form,

And, the required equivalent system is,

Augumented matrix:

Since

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

So, the system has one free variable.

let it be

Step 3

Now,

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