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:

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

Hint:

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

Answer 1

Given matrix \( A \) and the equation \( AX = B \), we need to solve for \( X \). The matrices are defined as:

\[A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}\]
The solution involves computing the inverse of \( A \), denoted \( A^{-1} \), and then calculating \( X = A^{-1}B \).

Step 1: Calculate the determinant of \( A \)
Using cofactor expansion along the first row:
\[\det(A) = 2 \cdot \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix}- (-3) \cdot \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix}+ 5 \cdot \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}\]
Evaluating the 2×2 determinants:
\[= 2(2 \cdot -2 - (-4) \cdot 1) + 3(3 \cdot -2 - (-4) \cdot 1) + 5(3 \cdot 1 - 2 \cdot 1) \\= 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1\]
Therefore, \( \det(A) = -1 \).

Step 2: Determine 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: Compute \( A^{-1} \)
\[A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = -1 \cdot \text{adj}(A) =\begin{bmatrix}0 & -11 & -2 \\-2 & 9 & -23 \\-1 & 5 & -13\end{bmatrix}\]

Step 4: Calculate \( X = A^{-1}B \)
\[X = A^{-1}B =\begin{bmatrix}0 & -11 & -2 \\-2 & 9 & -23 \\-1 & 5 & -13\end{bmatrix}\begin{bmatrix}11 \\ -5 \\ -3\end{bmatrix}\]
Performing the matrix multiplication:
\[x = 0 \cdot 11 + (-11) \cdot (-5) + (-2) \cdot (-3) = 55 + 6 = 61 \\y = -2 \cdot 11 + 9 \cdot (-5) + (-23) \cdot (-3) = -22 - 45 + 69 = 2 \\z = -1 \cdot 11 + 5 \cdot (-5) + (-13) \cdot (-3) = -11 - 25 + 39 = 3\]
Final Solution: \[x = 61, \quad y = 2, \quad z = 3\]