Question Solve the system of linear equations represented by the given augmented matrix using Gauss-Jordan method. Perform the row operations and find the values of $x$ , $y$ , and $z$ .

Studdy Solution

STEP 1

Assumptions
1. We are given an augmented matrix representing a system of linear equations.
2. We will use the Gauss-Jordan elimination method to solve the system.
3. The row operations are given step by step, and we need to apply them to the matrix.
4. The final matrix will be in reduced-row echelon form, from which we can read off the solutions for $x$ , $y$ , and $z$ .

STEP 2

The given augmented matrix is: $\left[\begin{array}{cccc} 2 & 1 & 1 & 6 \\ 1 & -1 & -1 & -3 \\ 3 & 1 & 2 & 7 \end{array}\right]$ The first row operation is $R_{1} \longleftrightarrow R_{2}$ , which means we swap the first and second rows.

STEP 3

After swapping $R_{1}$ and $R_{2}$ , the matrix becomes: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 2 & 1 & 1 & 6 \\ 3 & 1 & 2 & 7 \end{array}\right]$

STEP 4

The next row operations are $R_{2}=-2 R_{1}+R_{2}$ and $R_{3}=-3 R_{1}+R_{3}$ . Apply these operations to eliminate the coefficients below the leading 1 in the first column.

STEP 5

First, apply $R_{2}=-2 R_{1}+R_{2}$ : $R_{2} = -2 \cdot \left[\begin{array}{c} 1 \\ -1 \\ -1 \\ -3 \end{array}\right] + \left[\begin{array}{c} 2 \\ 1 \\ 1 \\ 6 \end{array}\right] = \left[\begin{array}{c} 0 \\ 3 \\ 3 \\ 12 \end{array}\right]$

STEP 6

Now, apply $R_{3}=-3 R_{1}+R_{3}$ : $R_{3} = -3 \cdot \left[\begin{array}{c} 1 \\ -1 \\ -1 \\ -3 \end{array}\right] + \left[\begin{array}{c} 3 \\ 1 \\ 2 \\ 7 \end{array}\right] = \left[\begin{array}{c} 0 \\ 4 \\ 5 \\ 14 \end{array}\right]$

STEP 7

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 3 & 3 & 12 \\ 0 & 4 & 5 & 14 \end{array}\right]$

STEP 8

The next operation is $R_{2}=\frac{1}{3} R_{2}$ , which scales the second row to create a leading 1 in the second column.

STEP 9

Apply the operation to $R_{2}$ : $R_{2} = \frac{1}{3} \cdot \left[\begin{array}{c} 0 \\ 3 \\ 3 \\ 12 \end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right]$

STEP 10

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & -1 & -1 & -3 \\ 0 & 1 & 1 & 4 \\ 0 & 4 & 5 & 14 \end{array}\right]$

STEP 11

The next operations are $R_{1}=R_{2}+R_{1}$ and $R_{3}=-4 R_{2}+R_{3}$ to create zeros below and above the leading 1 in the second column.

STEP 12

First, apply $R_{1}=R_{2}+R_{1}$ : $R_{1} = \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] + \left[\begin{array}{c} 1 \\ -1 \\ -1 \\ -3 \end{array}\right] = \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array}\right]$

STEP 13

Now, apply $R_{3}=-4 R_{2}+R_{3}$ : $R_{3} = -4 \cdot \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] + \left[\begin{array}{c} 0 \\ 4 \\ 5 \\ 14 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 2 \end{array}\right]$

STEP 14

The matrix now looks like this: $\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right]$

STEP 15

The last operation is $R_{2}=-R_{3}+R_{2}$ to create a zero above the leading 1 in the third column.

STEP 16

Apply the operation to $R_{2}$ : $R_{2} = - \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 2 \end{array}\right] + \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 4 \end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 2 \end{array}\right]$

STEP 17

The final matrix in reduced-row echelon form is: $\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]$

STEP 18

From the final matrix, we can read off the solutions for $x$ , $y$ , and $z$ directly as the values in the last column.

STEP 19

The solution for $x$ is: $x = 1$

STEP 20

The solution for $y$ is: $y = 2$

STEP 21

The solution for $z$ is: $z = 2$
The solutions for the variables are $x = 1$ , $y = 2$ , and $z = 2$ .

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