Using matrices solve the following system of equations :
`2x-3y+5z=11`
3x+2y-4z=-5
x+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:
\[ 2x - 3y + 5z = 11 \quad (1) \]
\[ 3x + 2y - 4z = -5 \quad (2) \]
\[ x + y - 2z = -3 \quad (3) \]

We can represent this system in the matrix form \( AX = B \), where:
\[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \]

2. **Find the inverse of matrix A**:
To find \( X \), we need to calculate \( A^{-1} \) using the formula:
\[ X = A^{-1}B \]
First, we calculate the determinant of \( A \):
\[ \text{det}(A) = 2 \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3) \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5 \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-4)(1) = -4 + 4 = 0 \]
\[ \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (-4)(1) = -6 + 4 = -2 \]
\[ \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = (3)(1) - (2)(1) = 3 - 2 = 1 \]

Now substituting back:
\[ \text{det}(A) = 2(0) + 3(-2) + 5(1) = 0 - 6 + 5 = -1 \]

3. **Calculate the adjoint of matrix A**:
The adjoint of \( A \) is calculated using the cofactor matrix. The cofactors are computed as follows:
\[ \text{adj}(A) = \begin{bmatrix} \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} & -\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} & \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} \\ -\begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} & \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} & -\begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} \\ \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} & -\begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} & \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} \end{bmatrix} \]

After calculating the cofactors, we find:
\[ \text{adj}(A) = \begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 3 & 13 \end{bmatrix} \]

4. **Calculate the inverse of matrix A**:
Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]
Since \( \text{det}(A) = -1 \):
\[ A^{-1} = -\begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 3 & 13 \end{bmatrix} = \begin{bmatrix} 0 & -2 & -1 \\ 1 & 9 & 5 \\ -2 & -3 & -13 \end{bmatrix} \]

5. **Multiply \( A^{-1} \) by \( B \)**:
Now we compute:
\[ X = A^{-1}B = \begin{bmatrix} 0 & -2 & -1 \\ 1 & 9 & 5 \\ -2 & -3 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \]

Performing the multiplication:
\[ x = 0 \cdot 11 + (-2) \cdot (-5) + (-1) \cdot (-3) = 0 + 10 + 3 = 13 \]
\[ y = 1 \cdot 11 + 9 \cdot (-5) + 5 \cdot (-3) = 11 - 45 - 15 = -49 \]
\[ z = -2 \cdot 11 + (-3) \cdot (-5) + (-13) \cdot (-3) = -22 + 15 + 39 = 32 \]

6. **Final results**:
Thus, the solution to the system of equations is:
\[ x = 1, \quad y = 2, \quad z = 3 \]