Question

If \(A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}\), find \(A^{-1}\) and use it to solve the following system of equations: \[ x - 2y = 10, \quad 2x - y - z = 8, \quad -2y + z = 7. \]

Hint:

To solve a system of linear equations using the inverse matrix method, express the system in the form \(A \cdot \vec{x} = \vec{b}\), compute \(A^{-1}\), and use \(\vec{x} = A^{-1} \cdot \vec{b}\).

Answer 1

Step 1: Express the system in matrix form.

The given system of equations can be represented as: \[ A \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}, \] 

where \[ A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}, \quad \text{and} \quad \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \text{ is the constant matrix.} \] 

Step 2: Calculate \(A^{-1}\). The inverse of a \(3 \times 3\) matrix \(A\) is computed using the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A), \] 

where \(\text{det}(A)\) is the determinant of \(A\) and \(\text{adj}(A)\) is the adjugate of \(A\). 

(a) Determine \(\text{det}(A)\): \[ \text{det}(A) = \begin{vmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{vmatrix}. \] 

Expanding along the first row yields: \[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 2 & -1 \\ 0 & -2 \end{vmatrix}. \] 

The minors are calculated as: \[ \begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} = (-1)(1) - (-1)(-2) = -1 - 2 = -3, \] \[ \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} = (2)(1) - (-1)(0) = 2 - 0 = 2. \] Substituting these values back: \[ \text{det}(A) = 1(-3) - (-2)(2) + 0 = -3 + 4 = 1. \] 

(b) Compute \(\text{adj}(A)\): The adjugate of \(A\) is the transpose of its cofactor matrix. 

The cofactors for each element of \(A\) are: \[ \text{Cofactor matrix of } A = \begin{bmatrix} -3 & 2 & 4 \\ 1 & 1 & -2 \\ 4 & 2 & 5 \end{bmatrix}. \] 
Therefore, the adjugate matrix is: \[ \text{adj}(A) = \begin{bmatrix} -3 & 1 & 4 \\ 2 & 1 & 2 \\ 4 & -2 & 5 \end{bmatrix}. \] 

(c) Calculate \(A^{-1}\): Since \(\text{det}(A) = 1\), the formula simplifies to: \[ A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \text{adj}(A). \] 

Thus: \[ A^{-1} = \begin{bmatrix} -3 & 1 & 4 \\ 2 & 1 & 2 \\ 4 & -2 & 5 \end{bmatrix}. \] Step 3: Solve for \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\). 

Using the relation: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} \cdot \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}, \] we compute the matrix product: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3 & 1 & 4 \\ 2 & 1 & 2 \\ 4 & -2 & 5 \end{bmatrix} \cdot \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}. \] 

Performing the multiplication: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3(10) + 1(8) + 4(7) \\ 2(10) + 1(8) + 2(7) \\ 4(10) - 2(8) + 5(7) \end{bmatrix}. \] 

Simplifying each element: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -30 + 8 + 28 \\ 20 + 8 + 14 \\ 40 - 16 + 35 \end{bmatrix} = \begin{bmatrix} 6 \\ 42 \\ 59 \end{bmatrix}. \] 

Final Solution: \[ x = 6, \quad y = 42, \quad z = 59. \]