c)
  {
  x - y - z = 0
  x + 2y - 5z = 2
  3x - 2y - 4z = 1
  }

To solve this system of equations, we can use methods such as substitution, elimination, or matrix operations. Here, I'll use matrix operations to find the solution.

The system of equations is:

\[
\begin{cases}
x - y - z = 0 \\
x + 2y - 5z = 2 \\
3x - 2y - 4z = 1
\end{cases}
\]

We can represent this system in matrix form \(Ax = b\), where \(A\) is the coefficient matrix, \(x\) is the vector of variables, and \(b\) is the constant vector:

\[
A = \begin{pmatrix}
1 & -1 & -1 \\
1 & 2 & -5 \\
3 & -2 & -4 
\end{pmatrix}, \quad
x = \begin{pmatrix}
x \\
y \\
z 
\end{pmatrix}, \quad
b = \begin{pmatrix}
0 \\
2 \\
1 
\end{pmatrix}
\]

To solve for \(x\), we can use the inverse of matrix \(A\):

\[
x = A^{-1}b
\]

First, we need to find the inverse of \(A\). Let's proceed with the calculations:

1. **Find the determinant of \(A\):**

\[
\text{det}(A) = 1 \cdot (2 \cdot (-4) - (-5) \cdot (-2)) - (-1) \cdot (1 \cdot (-4) - (-5) \cdot 3) + (-1) \cdot (1 \cdot (-2) - 2 \cdot 3)
\]

\[
\text{det}(A) = 1 \cdot (-8 - 10) - (-1) \cdot (-4 - 15) + (-1) \cdot (-2 - 6)
\]

\[
\text{det}(A) = 1 \cdot (-18) + 1 \cdot (-19) + 1 \cdot (-8)
\]

\[
\text{det}(A) = -18 - 19 - 8 = -45
\]

Since the determinant is not zero, the inverse exists.

2. **Find the inverse of \(A\), \(A^{-1}\):**

\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)
\]

Where \(\text{adj}(A)\) is the adjugate of \(A\). Calculate the minors and cofactors to find \(\text{adj}(A)\):

\[
\text{adj}(A) = \begin{pmatrix}
-14 & -11 & -4 \\
-7 & -11 & -5 \\
-4 & -9 & 3 
\end{pmatrix}
\]

\[
A^{-1} = \frac{1}{-45} \cdot \begin{pmatrix}
-14 & -11 & -4 \\
-7 & -11 & -5 \\
-4 & -9 & 3 
\end{pmatrix}
\]

\[
A^{-1} = \begin{pmatrix}
\frac{14}{45} & \frac{11}{45} & \frac{4}{45} \\
\frac{7}{45} & \frac{11}{45} & \frac{5}{45} \\
\frac{4}{45} & \frac{9}{45} & -\frac{3}{45} 
\end{pmatrix}
\]

3. **Multiply \(A^{-1}\) by \(b\):**

\[
x = A^{-1}b = \begin{pmatrix}
\frac{14}{45} & \frac{11}{45} & \frac{4}{45} \\
\frac{7}{45} & \frac{11}{45} & \frac{5}{45} \\
\frac{4}{45} & \frac{9}{45} & -\frac{3}{45} 
\end{pmatrix} \begin{pmatrix}
0 \\
2 \\
1 
\end{pmatrix}
\]

\[
x = \begin{pmatrix}
\frac{14 \cdot 0 + 11 \cdot 2 + 4 \cdot 1}{45} \\
\frac{7 \cdot 0 + 11 \cdot 2 + 5 \cdot 1}{45} \\
\frac{4 \cdot 0 + 9 \cdot 2 - 3 \cdot 1}{45}
\end{pmatrix}
= \begin{pmatrix}
\frac{22 + 4}{45} \\
\frac{22 + 5}{45} \\
\frac{18 - 3}{45}
\end{pmatrix}
= \begin{pmatrix}
\frac{26}{45} \\
\frac{27}{45} \\
\frac{15}{45}
\end{pmatrix}
= \begin{pmatrix}
\frac{26}{45} \\
\frac{27}{45} \\
\frac{1}{3}
\end{pmatrix}
\]

So, the solution is:

\[
x = \frac{26}{45}, \quad y = \frac{27}{45}, \quad z = \frac{1}{3}
\]

Would you like further details or explanations on any step? Here are some other questions you might find helpful:

1. How can I solve a system of equations using substitution?
2. What are the properties of determinants?
3. How can I use row reduction to solve a system of linear equations?
4. What is the significance of the adjugate matrix in finding the inverse?
5. How can I verify the solution of a system of equations?

**Tip:** When dealing with systems of linear equations, always check the determinant of the coefficient matrix to determine if the system has a unique solution.

Learn how to solve the system of linear equations x - y - z = 0, x + 2y - 5z = 2, 3x - 2y - 4z = 1 using matrix operations and inverse matrices. This problem covers key concepts including matrix representation of linear systems and the inverse matrix formula. Suitable for high school students.

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