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: \[ x = 6, \quad y = 42, \quad z = 59. \]
Let \( f(x) = \log x \) and \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \] Then the domain of \( f \circ g \) is: