Using matrices, solve the following system of equations for x,y and z.
`x+y+z=3`, `y+3z=4`, `x-2y+z=0`

Answer

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

1. **Write the system of equations in matrix form**:
The given equations are:
\[ \begin{align*} x + y + z &= 3 \quad \text{(1)} \\ y + 3z &= 4 \quad \text{(2)} \\ x - 2y + z &= 0 \quad \text{(3)} \end{align*} \]
We can express this in the form \( A \mathbf{x} = \mathbf{b} \), where:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \]

2. **Calculate the determinant \( \Delta \) of matrix \( A \)**:
\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{vmatrix} \]
To calculate the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
\[ \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = (1)(1) - (3)(-2) = 1 + 6 = 7 \]
\[ \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = (0)(1) - (3)(1) = -3 \]
\[ \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} = (0)(-2) - (1)(1) = -1 \]
Now substituting back:
\[ \Delta = 1 \cdot 7 - 1 \cdot (-3) + 1 \cdot (-1) = 7 + 3 - 1 = 9 \]

3. **Calculate \( \Delta_x \)**:
Replace the first column of \( A \) with \( \mathbf{b} \):
\[ \Delta_x = \begin{vmatrix} 3 & 1 & 1 \\ 4 & 1 & 3 \\ 0 & -2 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_x = 3 \cdot \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & 3 \\ 0 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 1 \\ 0 & -2 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 7 \quad \text{(calculated previously)} \]
\[ \begin{vmatrix} 4 & 3 \\ 0 & 1 \end{vmatrix} = 4 \]
\[ \begin{vmatrix} 4 & 1 \\ 0 & -2 \end{vmatrix} = -8 \]
Now substituting back:
\[ \Delta_x = 3 \cdot 7 - 1 \cdot 4 + 1 \cdot (-8) = 21 - 4 - 8 = 9 \]

4. **Calculate \( \Delta_y \)**:
Replace the second column of \( A \) with \( \mathbf{b} \):
\[ \Delta_y = \begin{vmatrix} 1 & 3 & 1 \\ 0 & 4 & 3 \\ 1 & 0 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_y = 1 \cdot \begin{vmatrix} 4 & 3 \\ 0 & 1 \end{vmatrix} - 3 \cdot \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \begin{vmatrix} 4 & 3 \\ 0 & 1 \end{vmatrix} = 4 \]
\[ \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = -3 \]
\[ \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = -4 \]
Now substituting back:
\[ \Delta_y = 1 \cdot 4 - 3 \cdot (-3) + 1 \cdot (-4) = 4 + 9 - 4 = 9 \]

5. **Calculate \( \Delta_z \)**:
Replace the third column of \( A \) with \( \mathbf{b} \):
\[ \Delta_z = \begin{vmatrix} 1 & 1 & 3 \\ 0 & 1 & 4 \\ 1 & -2 & 0 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_z = 1 \cdot \begin{vmatrix} 1 & 4 \\ -2 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ \begin{vmatrix} 1 & 4 \\ -2 & 0 \end{vmatrix} = 8 \]
\[ \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = -4 \]
\[ \begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix} = -1 \]
Now substituting back:
\[ \Delta_z = 1 \cdot 8 - 1 \cdot (-4) + 3 \cdot (-1) = 8 + 4 - 3 = 9 \]

6. **Calculate the values of \( x, y, z \)**:
Using Cramer's Rule:
\[ x = \frac{\Delta_x}{\Delta} = \frac{9}{9} = 1 \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{9}{9} = 1 \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{9}{9} = 1 \]

Thus, the solution to the system of equations is:
\[ x = 1, \quad y = 1, \quad z = 1 \]