Find `A^(-1)`, where `A=[{:(4,2,3),(1,1,1),(3,1,-2):}]` Hence, solve the following system equations
`{:(4x+2y+3z=2),(x+y+z=1),(3x+y-2z=5):}`

Answer

To find the inverse of the matrix \( A \) and solve the system of equations, we will follow these steps:

### Step 1: Write the Matrix \( A \)
The given matrix \( A \) is:
\[ A = \begin{pmatrix} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{pmatrix} \]

### Step 2: Calculate the Determinant of \( A \)
We will calculate the determinant \( \text{det}(A) \) using the formula for a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

For our matrix:
- \( a = 4, b = 2, c = 3 \)
- \( d = 1, e = 1, f = 1 \)
- \( g = 3, h = 1, i = -2 \)

Calculating the determinant:
\[ \text{det}(A) = 4(1 \cdot (-2) - 1 \cdot 1) - 2(1 \cdot (-2) - 1 \cdot 3) + 3(1 \cdot 1 - 1 \cdot 3) \]
\[ = 4(-2 - 1) - 2(-2 - 3) + 3(1 - 3) \]
\[ = 4(-3) - 2(-5) + 3(-2) \]
\[ = -12 + 10 - 6 = -8 \]

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

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} = (1)(-2) - (1)(1) = -2 - 1 = -3 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{pmatrix} 1 & 1 \\ 3 & -2 \end{pmatrix} = -((1)(-2) - (1)(3)) = -(-2 - 3) = 5 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det}\begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} = (1)(1) - (1)(3) = 1 - 3 = -2 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det}\begin{pmatrix} 2 & 3 \\ 1 & -2 \end{pmatrix} = -((2)(-2) - (3)(1)) = -(-4 - 3) = 7 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det}\begin{pmatrix} 4 & 3 \\ 3 & -2 \end{pmatrix} = (4)(-2) - (3)(3) = -8 - 9 = -17 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det}\begin{pmatrix} 4 & 2 \\ 3 & 1 \end{pmatrix} = -((4)(1) - (2)(3)) = -(4 - 6) = 2 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \text{det}\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} = (2)(1) - (3)(1) = 2 - 3 = -1 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{pmatrix} 4 & 3 \\ 1 & 1 \end{pmatrix} = -((4)(1) - (3)(1)) = -(4 - 3) = -1 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det}\begin{pmatrix} 4 & 2 \\ 1 & 1 \end{pmatrix} = (4)(1) - (2)(1) = 4 - 2 = 2 \]

Now, the cofactor matrix is:
\[ \text{Cofactor}(A) = \begin{pmatrix} -3 & 5 & -2 \\ 7 & -17 & 2 \\ -1 & -1 & 2 \end{pmatrix} \]

Taking the transpose gives us the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} -3 & 7 & -1 \\ 5 & -17 & -1 \\ -2 & 2 & 2 \end{pmatrix} \]

### Step 4: Calculate the Inverse of \( A \)
Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{-8} \begin{pmatrix} -3 & 7 & -1 \\ 5 & -17 & -1 \\ -2 & 2 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{8} & -\frac{7}{8} & \frac{1}{8} \\ -\frac{5}{8} & \frac{17}{8} & \frac{1}{8} \\ \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \end{pmatrix} \]

### Step 5: Solve the System of Equations
The system of equations can be written in matrix form \( AX = B \):
\[ X = A^{-1}B \]
where \( B = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \).

Calculating \( X \):
\[ X = \begin{pmatrix} \frac{3}{8} & -\frac{7}{8} & \frac{1}{8} \\ -\frac{5}{8} & \frac{17}{8} & \frac{1}{8} \\ \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} \]

Calculating each component:
1. \( x = \frac{3}{8}(2) - \frac{7}{8}(1) + \frac{1}{8}(5) = \frac{6}{8} - \frac{7}{8} + \frac{5}{8} = \frac{4}{8} = \frac{1}{2} \)
2. \( y = -\frac{5}{8}(2) + \frac{17}{8}(1) + \frac{1}{8}(5) = -\frac{10}{8} + \frac{17}{8} + \frac{5}{8} = \frac{12}{8} = \frac{3}{2} \)
3. \( z = \frac{1}{4}(2) - \frac{1}{4}(1) - \frac{1}{4}(5) = \frac{2}{4} - \frac{1}{4} - \frac{5}{4} = -\frac{4}{4} = -1 \)

Thus, the solution to the system of equations is:
\[ x = \frac{1}{2}, \quad y = \frac{3}{2}, \quad z = -1 \]