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: Represent the system in matrix form.

The given system of equations can be written 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: Find \( A^{-1} \).

The inverse of a \( 3 \times 3 \) matrix \( A \) is given by:

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

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

(a) Compute \(\text{det}(A)\):

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

Expanding along the first row:

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

Compute the minors:

\[ \begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} = (-1)(1) - (-1)(-2) = -1 - 2 = -3, \]

\[ \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} = (2)(1) - (-1)(0) = 2 - 0 = 2. \]

Substitute 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 the cofactor matrix. Compute the cofactors for each element of \( A \):

\[ \text{Cofactor matrix of } A = \begin{bmatrix} -3 & 2 & 4 \\ 1 & 1 & -2 \\ 4 & 2 & 5 \end{bmatrix}. \]

Thus:

\[ \text{adj}(A) = \begin{bmatrix} -3 & 1 & 4 \\ 2 & 1 & 2 \\ 4 & -2 & 5 \end{bmatrix}. \]

(c) Compute \( A^{-1} \):

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

as \(\text{det}(A) = 1\). 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 formula:

\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} \cdot \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}, \]

compute the product:

\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3 & 1 & 4 \\ 2 & 1 & 2 \\ 4 & -2 & 5 \end{bmatrix} \cdot \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}. \]

Perform the multiplication:

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

Simplify each term:

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

Final Answer:

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