Using matrices, solve the following system of equations for x,y and z.
`2x+3y+3z=5`, `x-2y+z=-4`, `3x-y-2z=3`

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:
\[ \begin{align*} 2x + 3y + 3z &= 5 \quad \text{(1)} \\ x - 2y + z &= -4 \quad \text{(2)} \\ 3x - y - 2z &= 3 \quad \text{(3)} \end{align*} \]
We can represent this in the form \(AX = B\), where:
\[ A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} \]

2. **Calculate the determinant of matrix A (\( \Delta \))**:
The determinant of \(A\) can be calculated using the formula for a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \(A\):
\[ \Delta = 2((-2)(-2) - (1)(-1)) - 3((1)(-2) - (1)(-2)) + 3((1)(-1) - (3)(-2)) \]
Calculating this gives:
\[ \Delta = 2(4 + 1) - 3(-2 + 2) + 3(-1 + 6) = 10 + 0 + 15 = 25 \]

3. **Calculate \( \Delta_x \)**:
Replace the first column of matrix \(A\) with matrix \(B\):
\[ \Delta_x = \begin{vmatrix} 5 & 3 & 3 \\ -4 & -2 & 1 \\ 3 & -1 & -2 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_x = 5((-2)(-2) - (1)(-1)) - 3((-4)(-2) - (1)(3)) + 3((-4)(-1) - (-2)(3)) \]
Simplifying gives:
\[ \Delta_x = 5(4 + 1) - 3(8 - 3) + 3(4 + 6) = 25 - 15 + 30 = 40 \]

4. **Calculate \( \Delta_y \)**:
Replace the second column of matrix \(A\) with matrix \(B\):
\[ \Delta_y = \begin{vmatrix} 2 & 5 & 3 \\ 1 & -4 & 1 \\ 3 & 3 & -2 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_y = 2((-4)(-2) - (1)(3)) - 5((1)(-2) - (1)(3)) + 3((1)(3) - (-4)(3)) \]
Simplifying gives:
\[ \Delta_y = 2(8 - 3) - 5(-2 - 3) + 3(3 + 12) = 10 + 25 + 45 = 80 \]

5. **Calculate \( \Delta_z \)**:
Replace the third column of matrix \(A\) with matrix \(B\):
\[ \Delta_z = \begin{vmatrix} 2 & 3 & 5 \\ 1 & -2 & -4 \\ 3 & -1 & 3 \end{vmatrix} \]
Calculating this determinant:
\[ \Delta_z = 2((-2)(3) - (-4)(-1)) - 3((1)(3) - (-4)(3)) + 5((1)(-1) - (-2)(3)) \]
Simplifying gives:
\[ \Delta_z = 2(-6 - 4) - 3(3 + 12) + 5(-1 + 6) = -20 - 45 + 25 = -40 \]

6. **Calculate the values of \(x\), \(y\), and \(z\)**:
Using Cramer's Rule:
\[ x = \frac{\Delta_x}{\Delta} = \frac{40}{25} = 1.6 \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{80}{25} = 3.2 \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{-40}{25} = -1.6 \]

Thus, the solution to the system of equations is:
\[ x = 1.6, \quad y = 3.2, \quad z = -1.6 \]