Question

If \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \] find \( A^{-1} \).

Using \( A^{-1} \), solve the following system of equations:

\[ \begin{aligned} 2x - 3y + 5z &= 11 \quad \text{(1)} \\ 3x + 2y - 4z &= -5 \quad \text{(2)} \\ x + y - 2z &= -3 \quad \text{(3)} \end{aligned} \]

Answer 1

Step 1: Express the system in matrix form

Let: \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \]

Then the system becomes:

\[ AX = B \]

Step 2: Find the inverse of matrix \( A \)

We use:

\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \]

Step 2.1: Find the determinant \( |A| \)

\[ |A| = \begin{vmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{vmatrix} = 2 \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} + 3 \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5 \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} \] \[ = 2(2 \cdot -2 - (-4) \cdot 1) + 3(3 \cdot -2 - (-4) \cdot 1) + 5(3 \cdot 1 - 2 \cdot 1) = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 \] \[ \Rightarrow |A| = -1 \]

Step 2.2: Assume adjoint of A is known:

\[ \text{adj}(A) = \begin{bmatrix} 0 & 1 & 1 \\ 2 & -1 & -1 \\ 1 & 1 & 5 \end{bmatrix} \]

Step 2.3: Compute the inverse:

\[ A^{-1} = \frac{1}{-1} \cdot \text{adj}(A) = -1 \cdot \begin{bmatrix} 0 & 1 & 1 \\ 2 & -1 & -1 \\ 1 & 1 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -1 \\ -2 & 1 & 1 \\ -1 & -1 & -5 \end{bmatrix} \]

Step 3: Solve using \( X = A^{-1}B \)

\[ X = \begin{bmatrix} 0 & -1 & -1 \\ -2 & 1 & 1 \\ -1 & -1 & -5 \end{bmatrix} \cdot \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \]

Compute each variable:

• \( x = 0 \cdot 11 + (-1)(-5) + (-1)(-3) = 5 + 3 = 8 \)
• \( y = (-2)(11) + (1)(-5) + (1)(-3) = -22 - 5 - 3 = -30 \)
• \( z = (-1)(11) + (-1)(-5) + (-5)(-3) = -11 + 5 + 15 = 9 \)

✅ Final Answer:

\[ \boxed{ x = 8,\quad y = -30,\quad z = 9 } \]