Question:

If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), find \( A^{-1} \) and hence solve the system of equations:

\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)

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To solve $AX = B$ using matrices, find $A^{-1}$ using cofactors, then compute $X = A^{-1}B$.
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Solution and Explanation

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: \[ x = 61, \quad y = 2, \quad z = 3 \]
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