Solve the following equations, using inverse of a matrix :
`{:(2x+3y+3z=5),(x-2y+z=4),(3x-y-2z=3):}`

Answer

To solve the system of equations using the inverse of a matrix, we will follow these steps:

### Step 1: Write the equations in matrix form
We have the equations:
1. \(2x + 3y + 3z = 5\)
2. \(x - 2y + z = 4\)
3. \(3x - y - 2z = 3\)

We can express this in the form \(Ax = b\), where:
- \(A = \begin{pmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{pmatrix}\)
- \(x = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
- \(b = \begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix}\)

### Step 2: Find the determinant of matrix \(A\)
To find the determinant of \(A\), we use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\).

Calculating the determinant:
\[ \text{det}(A) = 2((-2)(-2) - (1)(-1)) - 3(1(-2) - (1)(3)) + 3(1(-1) - (-2)(3)) \]
Calculating each part:
- \(= 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6)\)
- \(= 2(5) + 3(5) + 3(5)\)
- \(= 10 + 15 + 15 = 40\)

### Step 3: Find the adjoint of matrix \(A\)
The adjoint of \(A\) is the transpose of the cofactor matrix. We calculate the cofactor matrix \(C\):

1. For \(C_{11}\): \(C_{11} = (-2)(-2) - (1)(-1) = 4 + 1 = 5\)
2. For \(C_{12}\): \(C_{12} = -((1)(-2) - (1)(3)) = -(-2 - 3) = 5\)
3. For \(C_{13}\): \(C_{13} = (1)(-1) - (-2)(3) = -1 + 6 = 5\)
4. For \(C_{21}\): \(C_{21} = -((3)(-2) - (3)(-1)) = -(-6 + 3) = 3\)
5. For \(C_{22}\): \(C_{22} = (2)(-2) - (3)(3) = -4 - 9 = -13\)
6. For \(C_{23}\): \(C_{23} = -((2)(-1) - (3)(3)) = -(-2 - 9) = 11\)
7. For \(C_{31}\): \(C_{31} = (3)(-2) - (3)(-1) = -6 + 3 = -3\)
8. For \(C_{32}\): \(C_{32} = -((2)(-2) - (3)(1)) = -(-4 - 3) = 7\)
9. For \(C_{33}\): \(C_{33} = (2)(-1) - (3)(1) = -2 - 3 = -5\)

The cofactor matrix \(C\) is:
\[ C = \begin{pmatrix} 5 & 5 & 5 \\ 3 & -13 & 11 \\ -3 & 7 & -5 \end{pmatrix} \]

Now, the adjoint of \(A\) is the transpose of \(C\):
\[ \text{adj}(A) = C^T = \begin{pmatrix} 5 & 3 & -3 \\ 5 & -13 & 7 \\ 5 & 11 & -5 \end{pmatrix} \]

### Step 4: Calculate the inverse of matrix \(A\)
The inverse of \(A\) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{40} \begin{pmatrix} 5 & 3 & -3 \\ 5 & -13 & 7 \\ 5 & 11 & -5 \end{pmatrix} \]

### Step 5: Solve for \(x\)
Now we can find \(x\) by multiplying \(A^{-1}\) with \(b\):
\[ x = A^{-1}b = \frac{1}{40} \begin{pmatrix} 5 & 3 & -3 \\ 5 & -13 & 7 \\ 5 & 11 & -5 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix} \]

Calculating the multiplication:
1. First row: \(5 \cdot 5 + 3 \cdot 4 - 3 \cdot 3 = 25 + 12 - 9 = 28\)
2. Second row: \(5 \cdot 5 - 13 \cdot 4 + 7 \cdot 3 = 25 - 52 + 21 = -6\)
3. Third row: \(5 \cdot 5 + 11 \cdot 4 - 5 \cdot 3 = 25 + 44 - 15 = 54\)

So we have:
\[ x = \frac{1}{40} \begin{pmatrix} 28 \\ -6 \\ 54 \end{pmatrix} = \begin{pmatrix} \frac{28}{40} \\ \frac{-6}{40} \\ \frac{54}{40} \end{pmatrix} = \begin{pmatrix} \frac{7}{10} \\ \frac{-3}{20} \\ \frac{27}{20} \end{pmatrix} \]

### Final Solution
Thus, the solution to the system of equations is:
- \(x = \frac{7}{10}\)
- \(y = \frac{-3}{20}\)
- \(z = \frac{27}{20}\)