We are given matrix \( A \) and need to solve \( AX = B \), where:
\[
A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \quad
B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}
\]
To solve \( AX = B \), we first find \( A^{-1} \) and then compute \( X = A^{-1}B \).
Step 1: Compute the determinant of \( A \)
Use cofactor expansion along the first row:
\[
\det(A) = 2 \cdot
\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix}
- (-3) \cdot
\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix}
+ 5 \cdot
\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}
\]
Now compute each 2×2 determinant:
\[
= 2(2 \cdot -2 - (-4) \cdot 1) + 3(3 \cdot -2 - (-4) \cdot 1) + 5(3 \cdot 1 - 2 \cdot 1) \\
= 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1
\]
So, \( \det(A) = -1 \)
Step 2: Compute the adjoint of \( A \)
This is the transpose of the cofactor matrix:
\[
\text{Cof}(A) =
\begin{bmatrix}
0 & 2 & 1 \\
11 & -9 & -5 \\
2 & 23 & 13
\end{bmatrix}
\quad \Rightarrow \quad
\text{adj}(A) =
\begin{bmatrix}
0 & 11 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{bmatrix}
\]
Step 3: Find \( A^{-1} \)
\[
A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = -1 \cdot \text{adj}(A) =
\begin{bmatrix}
0 & -11 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{bmatrix}
\]
Step 4: Compute \( X = A^{-1}B \)
\[
X = A^{-1}B =
\begin{bmatrix}
0 & -11 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{bmatrix}
\begin{bmatrix}
11 \\ -5 \\ -3
\end{bmatrix}
\]
Now compute each component:
\[
x = 0 \cdot 11 + (-11) \cdot (-5) + (-2) \cdot (-3) = 55 + 6 = 61 \\
y = -2 \cdot 11 + 9 \cdot (-5) + (-23) \cdot (-3) = -22 - 45 + 69 = 2 \\
z = -1 \cdot 11 + 5 \cdot (-5) + (-13) \cdot (-3) = -11 - 25 + 39 = 3
\]
Final Answer:
\[
x = 61, \quad y = 2, \quad z = 3
\]