Solve the system of equations using matrices. Use the Gaussian elimination method with​ back-substitution.
left brace Start 3 By 1 Matrix 1st Row 1st Column nothing 2nd Row 1st Column nothing 3rd Row 1st Column nothing EndMatrix
 
w
minus
x
minus
y
minus
z
equals
8
4 w
minus
3 x
minus
y
plus
z
equals
7
3 w
plus
x
minus
3 y
minus
z
equals
26
negative w
plus
3 x
plus
2 y
plus
3 z
equals
negative 13
Question content area bottom
Part 1
The solution set is StartSet left parenthesis nothing comma nothing comma nothing comma nothing right parenthesis EndSet
  
  
  
  
. 
​(Simplify your​ answers.)

To solve the system of equations using Gaussian elimination and back-substitution, we first convert the system of equations into an augmented matrix and then proceed with row operations to bring the matrix to row-echelon form. Afterward, we use back-substitution to find the values of \( w, x, y, \) and \( z \).

### Step 1: Write the system of equations
The system of equations is:

\[
\begin{aligned}
1. & \quad w - x - y - z = 8 \\
2. & \quad 4w - 3x - y + z = 7 \\
3. & \quad 3w + x - 3y - z = 26 \\
4. & \quad -w + 3x + 2y + 3z = -13
\end{aligned}
\]

### Step 2: Set up the augmented matrix
We can write the system of equations as the following augmented matrix:

\[
\left[
\begin{array}{cccc|c}
1 & -1 & -1 & -1 & 8 \\
4 & -3 & -1 & 1 & 7 \\
3 & 1 & -3 & -1 & 26 \\
-1 & 3 & 2 & 3 & -13
\end{array}
\right]
\]

### Step 3: Perform Gaussian elimination
We will now use row operations to bring the matrix into upper triangular (row-echelon) form.

#### First pivot (make the first column 1):

The first element of the first row is already 1. So, no changes are needed for the first row.

#### Eliminate the entries below the first pivot:

We use row operations to eliminate the first entries in rows 2, 3, and 4.

- \( R_2 \rightarrow R_2 - 4R_1 \)
- \( R_3 \rightarrow R_3 - 3R_1 \)
- \( R_4 \rightarrow R_4 + R_1 \)

After these operations, the matrix becomes:

\[
\left[
\begin{array}{cccc|c}
1 & -1 & -1 & -1 & 8 \\
0 & 1 & 3 & 5 & -25 \\
0 & 4 & 0 & 2 & 2 \\
0 & 2 & 1 & 2 & -5
\end{array}
\right]
\]

#### Second pivot (make the second column 1):

We divide row 2 by 1 (it’s already 1, so no change is needed).

#### Eliminate the entries below the second pivot:

We eliminate the second entries in rows 3 and 4 using row operations:

- \( R_3 \rightarrow R_3 - 4R_2 \)
- \( R_4 \rightarrow

Solve the system of equations using matrices. Use the Gaussian elimination method with back-substitution.

w - x - y - z = 8
4w - 3x - y + z = 7
3w + x - 3y - z = 26
-w + 3x + 2y + 3z = -13

Learn how to solve a system of equations using the Gaussian elimination method with back-substitution. This example walks through solving a 4-variable system step-by-step, covering key linear algebra concepts such as row operations, augmented matrices, and back-substitution. Suitable for students in Grades 10-12.

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