Step 1 (Matrix Multiplication and Governing Law): [1 Mark]

To solve the system of equations using matrices, we first compute the product $QP$. Given:

$$P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}, \quad Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$$

The product $QP$ is calculated as:

$$QP = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 2+4+0 & -2+6-4 & 0+8-8 \\ -4+4+0 & 4+6-4 & 0+8-8 \\ 2-2+0 & -2-3+5 & 0-4+10 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I$$

Thus, $QP = 6I$, which implies $P^{-1} = \frac{1}{6}Q$.

Step 2 (Step-by-Step Mathematical Working): [2 Marks]

The system of equations can be written in the form $PX = B$, where:

$$P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$$

The solution is given by $X = P^{-1}B$. Substituting $P^{-1} = \frac{1}{6}Q$:

$$X = \frac{1}{6} Q B = \frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$$

Performing the multiplication:

$$X = \frac{1}{6} \begin{bmatrix} (2)(3) + (2)(17) + (-4)(7) \\ (-4)(3) + (2)(17) + (-4)(7) \\ (2)(3) + (-1)(17) + (5)(7) \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 6 + 34 - 28 \\ -12 + 34 - 28 \\ 6 - 17 + 35 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix}$$

Step 3 (Final Concluded Answer): [1 Mark]

Dividing each element by 6, we get:

$$X = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \implies x = 2, y = -1, z = 4$$

💡 EL10 Concept Insight: Think of the matrix $Q$ as a "