https://file.aiquickdraw.com/mathbot/file/1726670381376hitnx65l.pdf

To solve the system of linear equations:

\[
3x + 2y - z = 1 \tag{1}
\]
\[
x + 2y + 2z = 0 \tag{2}
\]
\[
2x + y + 3z = -1 \tag{3}
\]

We can approach this problem using either substitution, elimination, or matrix methods. Let's use the matrix method, particularly Gaussian elimination, to solve this system.

### Step 1: Set up the augmented matrix
The system of equations can be written as an augmented matrix:

\[
\begin{pmatrix}
3 & 2 & -1 & | & 1 \\
1 & 2 & 2 & | & 0 \\
2 & 1 & 3 & | & -1
\end{pmatrix}
\]

### Step 2: Perform row operations to get row echelon form

1. Make the pivot in the first column a 1 by swapping row 1 and row 2:

\[
\begin{pmatrix}
1 & 2 & 2 & | & 0 \\
3 & 2 & -1 & | & 1 \\
2 & 1 & 3 & | & -1
\end{pmatrix}
\]

2. Eliminate the entries below the pivot in the first column by replacing \( R_2 \) with \( R_2 - 3R_1 \) and \( R_3 \) with \( R_3 - 2R_1 \):

\[
\begin{pmatrix}
1 & 2 & 2 & | & 0 \\
0 & -4 & -7 & | & 1 \\
0 & -3 & -1 & | & -1
\end{pmatrix}
\]

3. Make the pivot in the second column a 1 by dividing \( R_2 \) by -4:

\[
\begin{pmatrix}
1 & 2 & 2 & | & 0 \\
0 & 1 & \frac{7}{4} & | & -\frac{1}{4} \\
0 & -3 & -1 & | & -

Solve the following linear system: 3x + 2y – z = 1; x + 2y + 2z = 0; 2x + y + 3z = -1

This page provides a step-by-step solution to the linear system of equations 3x + 2y - z = 1, x + 2y + 2z = 0, and 2x + y + 3z = -1. Using the Gaussian elimination method, learn how to transform the system into row echelon form and solve for the variables. Suitable for Grades 11-12 or early college students.

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