Solve the following equations, using inverse of a matrix :
`{:(x+2y-3z=-4),(2x+3y+2z=2),(3x-3y-4z=11):}`

Answer

To solve the system of equations using the inverse of a matrix, we start with the equations given:

1. \( x + 2y - 3z = -4 \)
2. \( 2x + 3y + 2z = 2 \)
3. \( 3x - 3y - 4z = 11 \)

We can express this system 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.

### Step 1: Set up the matrices

The coefficient matrix \( A \) and the constant matrix \( b \) are defined as follows:

\[ A = \begin{pmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad b = \begin{pmatrix} -4 \\ 2 \\ 11 \end{pmatrix} \]

### Step 2: Find the determinant of matrix \( A \)

To find the inverse of matrix \( A \), we first need to calculate its determinant \( |A| \).

\[ |A| = 1 \cdot (3 \cdot (-4) - 2 \cdot (-3)) - 2 \cdot (2 \cdot (-4) - 2 \cdot 3) - 3 \cdot (2 \cdot (-3) - 3 \cdot 3) \]

Calculating each term:

\[ |A| = 1 \cdot (-12 + 6) - 2 \cdot (-8 - 6) - 3 \cdot (-6 - 9) \]
\[ = 1 \cdot (-6) - 2 \cdot (-14) - 3 \cdot (-15) \]
\[ = -6 + 28 + 45 \]
\[ = 67 \]

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

Next, we need to find the adjoint of \( A \). The adjoint is the transpose of the cofactor matrix.

To find the cofactor matrix, we calculate the minors for each element of \( A \) and apply the checkerboard pattern of signs:

\[ \text{Cofactor}(A) = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix} \]

Calculating the cofactors:

1. \( C_{11} = \begin{vmatrix} 3 & 2 \\ -3 & -4 \end{vmatrix} = (3)(-4) - (2)(-3) = -12 + 6 = -6 \)
2. \( C_{12} = -\begin{vmatrix} 2 & 2 \\ 3 & -4 \end{vmatrix} = -((2)(-4) - (2)(3)) = -(-8 - 6) = 14 \)
3. \( C_{13} = \begin{vmatrix} 2 & 3 \\ 3 & -3 \end{vmatrix} = (2)(-3) - (3)(3) = -6 - 9 = -15 \)
4. \( C_{21} = -\begin{vmatrix} 2 & -3 \\ -3 & -4 \end{vmatrix} = -((2)(-4) - (-3)(-3)) = -(-8 - 9) = 17 \)
5. \( C_{22} = \begin{vmatrix} 1 & -3 \\ 3 & -4 \end{vmatrix} = (1)(-4) - (-3)(3) = -4 + 9 = 5 \)
6. \( C_{23} = -\begin{vmatrix} 1 & 2 \\ 3 & -3 \end{vmatrix} = -((1)(-3) - (2)(3)) = -(-3 - 6) = 9 \)
7. \( C_{31} = \begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} = (2)(2) - (2)(3) = 4 - 6 = -2 \)
8. \( C_{32} = -\begin{vmatrix} 1 & -3 \\ 2 & 2 \end{vmatrix} = -((1)(2) - (-3)(2)) = -2 + 6 = -4 \)
9. \( C_{33} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \)

Thus, the cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{pmatrix} -6 & 14 & -15 \\ 17 & 5 & 9 \\ -2 & -4 & -1 \end{pmatrix} \]

Now, we take the transpose to get the adjoint:

\[ \text{Adj}(A) = \begin{pmatrix} -6 & 17 & -2 \\ 14 & 5 & -4 \\ -15 & 9 & -1 \end{pmatrix} \]

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

The inverse of \( A \) is given by:

\[ A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A) = \frac{1}{67} \begin{pmatrix} -6 & 17 & -2 \\ 14 & 5 & -4 \\ -15 & 9 & -1 \end{pmatrix} \]

### Step 5: Solve for \( x \)

Now we can find \( x \) using the formula \( x = A^{-1}b \):

\[ x = \frac{1}{67} \begin{pmatrix} -6 & 17 & -2 \\ 14 & 5 & -4 \\ -15 & 9 & -1 \end{pmatrix} \begin{pmatrix} -4 \\ 2 \\ 11 \end{pmatrix} \]

Calculating the product:

1. First row: \( (-6)(-4) + (17)(2) + (-2)(11) = 24 + 34 - 22 = 36 \)
2. Second row: \( (14)(-4) + (5)(2) + (-4)(11) = -56 + 10 - 44 = -90 \)
3. Third row: \( (-15)(-4) + (9)(2) + (-1)(11) = 60 + 18 - 11 = 67 \)

Thus, we have:

\[ x = \frac{1}{67} \begin{pmatrix} 36 \\ -90 \\ 67 \end{pmatrix} \]

### Step 6: Final values of \( x, y, z \)

Calculating each variable:

\[ x = \frac{36}{67}, \quad y = \frac{-90}{67}, \quad z = \frac{67}{67} = 1 \]

### Conclusion

Thus, the solution to the system of equations is:

\[ x = 3, \quad y = -2, \quad z = 1 \]