Using matrices, solve the following system of equations for x,y and z.
`x+y-z=-1`, `3x+y-2z=3`, `x-y-z=-1`

Answer

To solve the system of equations using matrices, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + y - z = -1 \)
2. \( 3x + y - 2z = 3 \)
3. \( x - y - z = -1 \)

We can express these equations in the matrix form \( Ax = b \), where:
- \( A \) is the coefficient matrix,
- \( x \) is the column matrix of variables, and
- \( b \) is the column matrix of constants.

Thus, we have:
\[ A = \begin{pmatrix} 1 & 1 & -1 \\ 3 & 1 & -2 \\ 1 & -1 & -1 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad b = \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \) (denoted as \( \Delta \))
To find \( \Delta \), we calculate the determinant of matrix \( A \):
\[ \Delta = \begin{vmatrix} 1 & 1 & -1 \\ 3 & 1 & -2 \\ 1 & -1 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & -2 \\ -1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -2 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -2 \\ -1 & -1 \end{vmatrix} = (1)(-1) - (-2)(-1) = -1 - 2 = -3 \)

2. \( \begin{vmatrix} 3 & -2 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (-2)(1) = -3 + 2 = -1 \)

3. \( \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (1)(1) = -3 - 1 = -4 \)

Now substituting back:
\[ \Delta = 1(-3) - 1(-1) - 1(-4) = -3 + 1 + 4 = 2 \]

### Step 3: Calculate \( \Delta_x \), \( \Delta_y \), and \( \Delta_z \)
To find \( \Delta_x \), we replace the first column of \( A \) with \( b \):
\[ \Delta_x = \begin{vmatrix} -1 & 1 & -1 \\ 3 & 1 & -2 \\ -1 & -1 & -1 \end{vmatrix} \]
Calculating \( \Delta_x \):
\[ \Delta_x = -1 \cdot \begin{vmatrix} 1 & -2 \\ -1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -2 \\ -1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 1 \\ -1 & -1 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. Already calculated as \(-3\).
2. \( \begin{vmatrix} 3 & -2 \\ -1 & -1 \end{vmatrix} = (3)(-1) - (-2)(-1) = -3 - 2 = -5 \)
3. \( \begin{vmatrix} 3 & 1 \\ -1 & -1 \end{vmatrix} = (3)(-1) - (1)(-1) = -3 + 1 = -2 \)

Now substituting back:
\[ \Delta_x = -1(-3) - 1(-5) - 1(-2) = 3 + 5 + 2 = 10 \]

Now, for \( \Delta_y \), we replace the second column of \( A \) with \( b \):
\[ \Delta_y = \begin{vmatrix} 1 & -1 & -1 \\ 3 & 3 & -2 \\ 1 & -1 & -1 \end{vmatrix} \]
Calculating \( \Delta_y \):
1. \( \begin{vmatrix} -1 & -1 \\ -1 & -1 \end{vmatrix} = 0 \)
2. \( \begin{vmatrix} 3 & -2 \\ 1 & -1 \end{vmatrix} = -1 \)
3. \( \begin{vmatrix} 3 & 3 \\ 1 & -1 \end{vmatrix} = -6 \)

Now substituting back:
\[ \Delta_y = 1(0) - 3(-1) + 1(-6) = 0 + 3 - 6 = -3 \]

Finally, for \( \Delta_z \), we replace the third column of \( A \) with \( b \):
\[ \Delta_z = \begin{vmatrix} 1 & 1 & -1 \\ 3 & 1 & 3 \\ 1 & -1 & -1 \end{vmatrix} \]
Calculating \( \Delta_z \):
1. \( \begin{vmatrix} 1 & 3 \\ -1 & -1 \end{vmatrix} = 1 \)
2. \( \begin{vmatrix} 3 & 3 \\ 1 & -1 \end{vmatrix} = -6 \)
3. \( \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = -4 \)

Now substituting back:
\[ \Delta_z = 1(1) - 3(-6) + 1(-4) = 1 + 18 - 4 = 15 \]

### Step 4: Solve for \( x, y, z \)
Using Cramer’s rule:
\[ x = \frac{\Delta_x}{\Delta} = \frac{10}{2} = 5 \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{-3}{2} = -\frac{3}{2} \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{15}{2} = 7.5 \]

### Final Result
Thus, the solution is:
\[ x = 5, \quad y = -\frac{3}{2}, \quad z = 7.5 \]