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The image shows a system of linear equations represented both in standard form and in matrix form. Let's solve it step-by-step using matrix methods.

### System of Linear Equations:
From the image, the system of linear equations is given by:

\[
\begin{cases}
2x + y - z = 0 \\
x - y + z = -1
\end{cases}
\]

### Matrix Form:
The system can be represented in matrix form as:

\[
\begin{bmatrix}
2 & 1 & -1 \\
1 & -1 & 1 
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z 
\end{bmatrix}
=
\begin{bmatrix}
0 \\
-1
\end{bmatrix}
\]

Let's denote the coefficient matrix by \(A\), the variable matrix by \(X\), and the constant matrix by \(B\):

\[
A = \begin{bmatrix}
2 & 1 & -1 \\
1 & -1 & 1 
\end{bmatrix}, \quad
X = \begin{bmatrix}
x \\
y \\
z 
\end{bmatrix}, \quad
B = \begin{bmatrix}
0 \\
-1
\end{bmatrix}
\]

### Solution Using Matrix Operations:
To solve for \(X\), we need to solve the matrix equation \(AX = B\). Since \(A\) is a \(2 \times 3\) matrix, we notice that there are fewer equations than variables, so this system might have an infinite number of solutions or be underdetermined.

#### Step 1: Write the Augmented Matrix
We write the augmented matrix \([A|B]\):

\[
\left[ \begin{array}{ccc|c}
2 & 1 & -1 & 0 \\
1 & -1 & 1 & -1 
\end{array} \right]
\]

#### Step 2: Row Reduction
Let's perform row operations to reduce this matrix to row echelon form.

1. Divide the first row by 2:
   \[
   \left[ \begin{array}{ccc|c}
   1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
   1 & -1 & 1 & -1 
   \end{array} \right]
   \]

2. Subtract the first row from the second row:
   \[
   \left[ \begin{array}{ccc|c}
   1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
   0 & -\frac{3}{2} & \frac{3}{2} & -1 
   \end{array} \right]
   \]

3. Multiply the second row by \(-\frac{2}{3}\):
   \[
   \left[ \begin{array}{ccc|c}
   1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
   0 & 1 & -1 & \frac{2}{3} 
   \end{array} \right]
   \]

4. Subtract \(\frac{1}{2}\) times the second row from the first row:
   \[
   \left[ \begin{array}{ccc|c}
   1 & 0 & 0 & -\frac{1}{3} \\
   0 & 1 & -1 & \frac{2}{3} 
   \end{array} \right]
   \]

#### Step 3: Interpretation of the Result

From the reduced matrix, we can interpret the equations as:

\[
x = -\frac{1}{3}, \quad y - z = \frac{2}{3}.
\]

So, \(y\) and \(z\) are free variables. Let's set \(z = t\), where \(t\) is a parameter. Then \(y = \frac{2}{3} + t\).

#### Final Solution:

\[
x = -\frac{1}{3}, \quad y = \frac{2}{3} + t, \quad z = t.
\]

Thus, the solution set is:

\[
\left( x, y, z \right) = \left( -\frac{1}{3}, \frac{2}{3} + t, t \right) \quad \text{for any } t \in \mathbb{R}.
\]

This system has infinitely many solutions along a line in three-dimensional space.

Would you like more details, or do you have any questions?

### Related Questions:

1. How do you interpret a matrix when there are more variables than equations?
2. What are the conditions for a unique solution in a system of linear equations?
3. How does the parameter \(t\) change the set of solutions in this context?
4. What is the geometric representation of this solution set in three dimensions?
5. How can we use matrix inverses to solve systems of equations when \(A\) is a square matrix?

### Tip:
When dealing with a system of linear equations, always check the dimensions of your coefficient matrix to understand the nature of the solution set (unique, infinite, or none).

Learn how to solve a system of linear equations using matrices with a step-by-step guide. This problem involves matrix operations and is suitable for high school students.

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