1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + y + z = 2 \\ 3x - 4y - 4z = -1 \\ 2x - 5y + 2z = -17 \end{cases}$$ 2. **Choose a method:** We will use **Cramer's Rule** which involves determinants of matrices. 3. **Write the coefficient matrix $A$ and constant vector $\mathbf{b}$:** $$A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & -4 & -4 \\ 2 & -5 & 2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 2 \\ -1 \\ -17 \end{bmatrix}$$ 4. **Calculate the determinant of $A$, $\det(A)$:** $$\det(A) = 1 \times \begin{vmatrix} -4 & -4 \\ -5 & 2 \end{vmatrix} - 1 \times \begin{vmatrix} 3 & -4 \\ 2 & 2 \end{vmatrix} + 1 \times \begin{vmatrix} 3 & -4 \\ 2 & -5 \end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix} -4 & -4 \\ -5 & 2 \end{vmatrix} = (-4)(2) - (-4)(-5) = -8 - 20 = -28$$ $$\begin{vmatrix} 3 & -4 \\ 2 & 2 \end{vmatrix} = 3 \times 2 - (-4) \times 2 = 6 + 8 = 14$$ $$\begin{vmatrix} 3 & -4 \\ 2 & -5 \end{vmatrix} = 3 \times (-5) - (-4) \times 2 = -15 + 8 = -7$$ So, $$\det(A) = 1 \times (-28) - 1 \times 14 + 1 \times (-7) = -28 - 14 - 7 = -49$$ 5. **Calculate determinants for $x$, $y$, and $z$ replacing respective columns with $\mathbf{b}$:** - For $x$, replace first column: $$A_x = \begin{bmatrix} 2 & 1 & 1 \\ -1 & -4 & -4 \\ -17 & -5 & 2 \end{bmatrix}$$ $$\det(A_x) = 2 \times \begin{vmatrix} -4 & -4 \\ -5 & 2 \end{vmatrix} - 1 \times \begin{vmatrix} -1 & -4 \\ -17 & 2 \end{vmatrix} + 1 \times \begin{vmatrix} -1 & -4 \\ -17 & -5 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -4 & -4 \\ -5 & 2 \end{vmatrix} = -28$$ (from before) $$\begin{vmatrix} -1 & -4 \\ -17 & 2 \end{vmatrix} = (-1)(2) - (-4)(-17) = -2 - 68 = -70$$ $$\begin{vmatrix} -1 & -4 \\ -17 & -5 \end{vmatrix} = (-1)(-5) - (-4)(-17) = 5 - 68 = -63$$ So, $$\det(A_x) = 2 \times (-28) - 1 \times (-70) + 1 \times (-63) = -56 + 70 - 63 = -49$$ - For $y$, replace second column: $$A_y = \begin{bmatrix} 1 & 2 & 1 \\ 3 & -1 & -4 \\ 2 & -17 & 2 \end{bmatrix}$$ $$\det(A_y) = 1 \times \begin{vmatrix} -1 & -4 \\ -17 & 2 \end{vmatrix} - 2 \times \begin{vmatrix} 3 & -4 \\ 2 & 2 \end{vmatrix} + 1 \times \begin{vmatrix} 3 & -1 \\ 2 & -17 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -1 & -4 \\ -17 & 2 \end{vmatrix} = -70$$ (from above) $$\begin{vmatrix} 3 & -4 \\ 2 & 2 \end{vmatrix} = 14$$ (from before) $$\begin{vmatrix} 3 & -1 \\ 2 & -17 \end{vmatrix} = 3 \times (-17) - (-1) \times 2 = -51 + 2 = -49$$ So, $$\det(A_y) = 1 \times (-70) - 2 \times 14 + 1 \times (-49) = -70 - 28 - 49 = -147$$ - For $z$, replace third column: $$A_z = \begin{bmatrix} 1 & 1 & 2 \\ 3 & -4 & -1 \\ 2 & -5 & -17 \end{bmatrix}$$ $$\det(A_z) = 1 \times \begin{vmatrix} -4 & -1 \\ -5 & -17 \end{vmatrix} - 1 \times \begin{vmatrix} 3 & -1 \\ 2 & -17 \end{vmatrix} + 2 \times \begin{vmatrix} 3 & -4 \\ 2 & -5 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -4 & -1 \\ -5 & -17 \end{vmatrix} = (-4)(-17) - (-1)(-5) = 68 - 5 = 63$$ $$\begin{vmatrix} 3 & -1 \\ 2 & -17 \end{vmatrix} = -49$$ (from above) $$\begin{vmatrix} 3 & -4 \\ 2 & -5 \end{vmatrix} = -7$$ (from before) So, $$\det(A_z) = 1 \times 63 - 1 \times (-49) + 2 \times (-7) = 63 + 49 - 14 = 98$$ 6. **Calculate the variables using Cramer's Rule:** $$x = \frac{\det(A_x)}{\det(A)} = \frac{-49}{-49} = 1$$ $$y = \frac{\det(A_y)}{\det(A)} = \frac{-147}{-49} = 3$$ $$z = \frac{\det(A_z)}{\det(A)} = \frac{98}{-49} = -2$$ **Final answer:** $$\boxed{x=1, y=3, z=-2}$$