Solve the following system of equation by matrix method
` x+2y +3z+1= 0, -2x+y-z +5=0 ,3x +y-2z -3=0`

Answer

To solve the system of equations using the matrix method, we start by rewriting the equations in a standard form:

1. \( x + 2y + 3z + 1 = 0 \) can be rewritten as:
\[ x + 2y + 3z = -1 \]

2. \( -2x + y - z + 5 = 0 \) can be rewritten as:
\[ -2x + y - z = -5 \]

3. \( 3x + y - 2z - 3 = 0 \) can be rewritten as:
\[ 3x + y - 2z = 3 \]

Now we have the following system of equations:
\[ \begin{align*} 1. & \quad x + 2y + 3z = -1 \\ 2. & \quad -2x + y - z = -5 \\ 3. & \quad 3x + y - 2z = 3 \end{align*} \]

### Step 1: Form the coefficient matrix \( A \), variable matrix \( X \), and constant matrix \( B \)

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ -2 & 1 & -1 \\ 3 & 1 & -2 \end{bmatrix} \]

The variable matrix \( X \) is:
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

The constant matrix \( B \) is:
\[ B = \begin{bmatrix} -1 \\ -5 \\ 3 \end{bmatrix} \]

### Step 2: Write the matrix equation

We can express the system of equations in matrix form as:
\[ AX = B \]

### Step 3: Find the inverse of matrix \( A \)

To find \( X \), we need to calculate \( A^{-1} \) and then use the formula:
\[ X = A^{-1}B \]

First, we calculate the determinant of \( A \):
\[ \text{det}(A) = 1 \cdot (1 \cdot (-2) - (-1) \cdot 1) - 2 \cdot (-2 \cdot (-2) - (-1) \cdot 3) + 3 \cdot (-2 \cdot 1 - 1 \cdot 3) \]
Calculating this step by step:
\[ = 1 \cdot (-2 + 1) - 2 \cdot (4 + 3) + 3 \cdot (-2 - 3) \]
\[ = 1 \cdot (-1) - 2 \cdot 7 + 3 \cdot (-5) \]
\[ = -1 - 14 - 15 = -30 \]

Next, we find the adjoint of \( A \). The adjoint is calculated using the cofactor matrix and then transposing it.

The cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} \text{det} \begin{bmatrix} 1 & -1 \\ 1 & -2 \end{bmatrix} & -\text{det} \begin{bmatrix} -2 & -1 \\ 3 & -2 \end{bmatrix} & \text{det} \begin{bmatrix} -2 & 1 \\ 3 & 1 \end{bmatrix} \\ -\text{det} \begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix} & \text{det} \begin{bmatrix} 1 & 3 \\ 3 & -2 \end{bmatrix} & -\text{det} \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} \\ \text{det} \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} & -\text{det} \begin{bmatrix} 1 & 3 \\ -2 & -1 \end{bmatrix} & \text{det} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \end{bmatrix} \]

Calculating the determinants:
1. \( \text{det} \begin{bmatrix} 1 & -1 \\ 1 & -2 \end{bmatrix} = 1 \cdot (-2) - (-1) \cdot 1 = -2 + 1 = -1 \)
2. \( -\text{det} \begin{bmatrix} -2 & -1 \\ 3 & -2 \end{bmatrix} = -((-2)(-2) - (-1)(3)) = -(4 + 3) = -7 \)
3. \( \text{det} \begin{bmatrix} -2 & 1 \\ 3 & 1 \end{bmatrix} = (-2)(1) - (1)(3) = -2 - 3 = -5 \)
4. \( -\text{det} \begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix} = -((2)(-2) - (3)(1)) = -(-4 - 3) = 7 \)
5. \( \text{det} \begin{bmatrix} 1 & 3 \\ 3 & -2 \end{bmatrix} = (1)(-2) - (3)(3) = -2 - 9 = -11 \)
6. \( -\text{det} \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} = -((1)(1) - (2)(3)) = -(1 - 6) = 5 \)
7. \( \text{det} \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} = (2)(-1) - (3)(1) = -2 - 3 = -5 \)
8. \( -\text{det} \begin{bmatrix} 1 & 3 \\ -2 & -1 \end{bmatrix} = -((1)(-1) - (3)(-2)) = -(-1 + 6) = -5 \)
9. \( \text{det} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} = (1)(1) - (2)(-2) = 1 + 4 = 5 \)

Putting it all together, the cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} -1 & -7 & -5 \\ 7 & -11 & 5 \\ -5 & -5 & 5 \end{bmatrix} \]

Now, we take the transpose of \( C \) to get the adjoint \( \text{adj}(A) \):
\[ \text{adj}(A) = \begin{bmatrix} -1 & 7 & -5 \\ -7 & -11 & -5 \\ -5 & 5 & 5 \end{bmatrix} \]

### Step 4: Calculate the inverse of \( A \)

Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{-30} \begin{bmatrix} -1 & 7 & -5 \\ -7 & -11 & -5 \\ -5 & 5 & 5 \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{1}{30} & -\frac{7}{30} & \frac{1}{6} \\ \frac{7}{30} & \frac{11}{30} & \frac{1}{6} \\ \frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} \end{bmatrix} \]

### Step 5: Multiply \( A^{-1} \) by \( B \)

Now we compute \( X = A^{-1}B \):
\[ X = \begin{bmatrix} \frac{1}{30} & -\frac{7}{30} & \frac{1}{6} \\ \frac{7}{30} & \frac{11}{30} & \frac{1}{6} \\ \frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} \end{bmatrix} \begin{bmatrix} -1 \\ -5 \\ 3 \end{bmatrix} \]

Calculating this gives:
\[ X = \begin{bmatrix} \frac{1}{30}(-1) + -\frac{7}{30}(-5) + \frac{1}{6}(3) \\ \frac{7}{30}(-1) + \frac{11}{30}(-5) + \frac{1}{6}(3) \\ \frac{1}{6}(-1) + -\frac{1}{6}(-5) + -\frac{1}{6}(3) \end{bmatrix} \]

Calculating each component:
1. First row:
\[ = \frac{-1}{30} + \frac{35}{30} + \frac{5}{30} = \frac{39}{30} = \frac{13}{10} \]
2. Second row:
\[ = \frac{-7}{30} - \frac{55}{30} + \frac{5}{30} = \frac{-57}{30} = -\frac{19}{10} \]
3. Third row:
\[ = \frac{-1}{6} + \frac{5}{6} - \frac{3}{6} = \frac{1}{6} \]

Thus, we have:
\[ X = \begin{bmatrix} \frac{13}{10} \\ -\frac{19}{10} \\ \frac{1}{6} \end{bmatrix} \]

### Final Result

The solution to the system of equations is:
\[ x = \frac{13}{10}, \quad y = -\frac{19}{10}, \quad z = \frac{1}{6} \]