Using matrices, solve the following system of equations for x,y and z.
`2x+y+z=1`, `x-2y-z=(3)/(2)`, `3y-5z=9`.

Answer

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

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( 2x + y + z = 1 \)
2. \( x - 2y - z = \frac{3}{2} \)
3. \( 3y - 5z = 9 \)

We can express this in the matrix form \( A \mathbf{x} = \mathbf{b} \), where:
- \( A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{pmatrix} \)
- \( \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \)
- \( \mathbf{b} = \begin{pmatrix} 1 \\ \frac{3}{2} \\ 9 \end{pmatrix} \)

### Step 2: Calculate the determinant of matrix A (\( \Delta \))
To find the determinant \( \Delta \) of matrix \( A \):
\[ \Delta = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta = 2 \begin{vmatrix} -2 & -1 \\ 3 & -5 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 0 & -5 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 0 & 3 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \Delta = 2((-2)(-5) - (-1)(3)) - 1(1 \cdot -5 - 0 \cdot -1) + 1(1 \cdot 3 - 0 \cdot -2) \]
\[ = 2(10 + 3) - (-5) + 3 = 2(13) + 5 + 3 = 26 + 5 + 3 = 34 \]

### Step 3: Calculate \( \Delta_x \)
Replace the first column of \( A \) with \( \mathbf{b} \) to find \( \Delta_x \):
\[ \Delta_x = \begin{vmatrix} 1 & 1 & 1 \\ \frac{3}{2} & -2 & -1 \\ 9 & 3 & -5 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_x = 1 \begin{vmatrix} -2 & -1 \\ 3 & -5 \end{vmatrix} - 1 \begin{vmatrix} \frac{3}{2} & -1 \\ 9 & -5 \end{vmatrix} + 1 \begin{vmatrix} \frac{3}{2} & -2 \\ 9 & 3 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \Delta_x = 1((-2)(-5) - (-1)(3)) - 1\left(\frac{3}{2}(-5) - (-1)(9)\right) + 1\left(\frac{3}{2}(3) - (-2)(9)\right) \]
\[ = 1(10 + 3) - 1\left(-\frac{15}{2} + 9\right) + 1\left(\frac{9}{2} + 18\right) \]
\[ = 13 + \frac{15}{2} - \frac{9}{2} - 18 = 13 + \frac{6}{2} - 18 = 13 + 3 - 18 = -2 \]

### Step 4: Calculate \( \Delta_y \)
Replace the second column of \( A \) with \( \mathbf{b} \) to find \( \Delta_y \):
\[ \Delta_y = \begin{vmatrix} 2 & 1 & 1 \\ 1 & \frac{3}{2} & -1 \\ 0 & 9 & -5 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_y = 2 \begin{vmatrix} \frac{3}{2} & -1 \\ 9 & -5 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 0 & -5 \end{vmatrix} + 1 \begin{vmatrix} 1 & \frac{3}{2} \\ 0 & 9 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \Delta_y = 2\left(\frac{3}{2}(-5) - (-1)(9)\right) - (-5) + 9 \]
\[ = 2\left(-\frac{15}{2} + 9\right) + 5 + 9 = 2\left(-\frac{15}{2} + \frac{18}{2}\right) + 14 = 2\left(\frac{3}{2}\right) + 14 = 3 + 14 = 17 \]

### Step 5: Calculate \( \Delta_z \)
Replace the third column of \( A \) with \( \mathbf{b} \) to find \( \Delta_z \):
\[ \Delta_z = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -2 & \frac{3}{2} \\ 0 & 3 & 9 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_z = 2 \begin{vmatrix} -2 & \frac{3}{2} \\ 3 & 9 \end{vmatrix} - 1 \begin{vmatrix} 1 & \frac{3}{2} \\ 0 & 9 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 0 & 3 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \Delta_z = 2((-2)(9) - (3)(\frac{3}{2})) - (1 \cdot 9) + 1(3) \]
\[ = 2(-18 - \frac{9}{2}) - 9 + 3 = 2\left(-\frac{36}{2} - \frac{9}{2}\right) - 6 = 2\left(-\frac{45}{2}\right) - 6 \]
\[ = -45 - 6 = -51 \]

### Step 6: Calculate the values of \( x, y, z \)
Using Cramer's Rule:
\[ x = \frac{\Delta_x}{\Delta} = \frac{-2}{34} = -\frac{1}{17} \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{17}{34} = \frac{1}{2} \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{-51}{34} = -\frac{3}{2} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = -\frac{1}{17}, \quad y = \frac{1}{2}, \quad z = -\frac{3}{2} \]