1. **State the problem:** Solve the system of equations using Cramer's Rule: $$\begin{cases} x - y + 4z = 0 \\ x - 2y = 1 \\ 2x - 5y - 4z = 2 \end{cases}$$ 2. **Write the coefficient matrix $A$ and constants vector $\mathbf{b}$:** $$A = \begin{bmatrix} 1 & -1 & 4 \\ 1 & -2 & 0 \\ 2 & -5 & -4 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$$ 3. **Calculate the determinant of $A$, $\det(A)$:** $$\det(A) = 1 \cdot \begin{vmatrix} -2 & 0 \\ -5 & -4 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & 0 \\ 2 & -4 \end{vmatrix} + 4 \cdot \begin{vmatrix} 1 & -2 \\ 2 & -5 \end{vmatrix}$$ Calculate each minor: $$\begin{aligned} M_{11} &= (-2)(-4) - (0)(-5) = 8 \\ M_{12} &= 1 \cdot (-4) - 0 \cdot 2 = -4 \\ M_{13} &= 1 \cdot (-5) - (-2) \cdot 2 = -5 + 4 = -1 \end{aligned}$$ So, $$\det(A) = 1 \cdot 8 - (-1) \cdot (-4) + 4 \cdot (-1) = 8 - 4 - 4 = 0$$ 4. Since $\det(A) = 0$, the system does not have a unique solution, so Cramer's Rule cannot be applied. **Final answer:** The determinant of the coefficient matrix is zero, so the system has either no solution or infinitely many solutions; Cramer's Rule does not apply.