Determine the value(s) of \( x \) such that \( \left[\begin{array}{lll}x & 2 & 1\end{array}\right]\left[\begin{array}{ccc}-4 & -1 & 4 \\ -1 & 2 & 3 \\ 4 & 11 & -1\end{array}\right]\left[\begin{array}{c}x \\ -1 \\ 4\end{array}\right]=[0] \) \( x= \) Note: If there is more than one value separate them by commas.

To solve the equation \[ \left[\begin{array}{lll}x & 2 & 1\end{array}\right]\left[\begin{array}{ccc}-4 & -1 & 4 \\ -1 & 2 & 3 \\ 4 & 11 & -1\end{array}\right]\left[\begin{array}{c}x \\ -1 \\ 4\end{array}\right]=[0], \] we will first compute the product of the matrices involved. Let us denote \[ A = \left[\begin{array}{ccc}-4 & -1 & 4 \\ -1 & 2 & 3 \\ 4 & 11 & -1\end{array}\right] \] and \[ \mathbf{v} = \left[\begin{array}{c}x \\ -1 \\ 4\end{array}\right]. \] First, we compute \( A \mathbf{v} \): \[ A \mathbf{v} = \left[\begin{array}{ccc}-4 & -1 & 4 \\ -1 & 2 & 3 \\ 4 & 11 & -1\end{array}\right]\left[\begin{array}{c}x \\ -1 \\ 4\end{array}\right] \] Calculating each component: 1st component: \[ -4x - 1(-1) + 4 \cdot 4 = -4x + 1 + 16 = -4x + 17 \] 2nd component: \[ -1 \cdot x + 2(-1) + 3 \cdot 4 = -x - 2 + 12 = -x + 10 \] 3rd component: \[ 4x + 11(-1) - 1 \cdot 4 = 4x - 11 - 4 = 4x - 15 \] Hence, \[ A \mathbf{v} = \left[\begin{array}{c}-4x + 17 \\ -x + 10 \\ 4x - 15\end{array}\right]. \] Next, we compute: \[ \left[\begin{array}{lll}x & 2 & 1\end{array}\right] \left[\begin{array}{c}-4x + 17 \\ -x + 10 \\ 4x - 15\end{array}\right]. \] This results in: \[ x(-4x + 17) + 2(-x + 10) + 1(4x - 15) = -4x^2 + 17x - 2x + 20 + 4x - 15. \] Combine like terms: \[ -4x^2 + (17x - 2x + 4x) + (20 - 15) = -4x^2 + 19x + 5. \] We need to set this expression to equal zero: \[ -4x^2 + 19x + 5 = 0. \] Multiplying through by -1 gives: \[ 4x^2 - 19x - 5 = 0. \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 4 \), \( b = -19 \), and \( c = -5 \): Calculate the discriminant: \[ b^2 - 4ac = (-19)^2 - 4(4)(-5) = 361 + 80 = 441. \] Now, substitute into the quadratic formula: \[ x = \frac{19 \pm \sqrt{441}}{8} = \frac{19 \pm 21}{8}. \] This gives two solutions: 1. \( x = \frac{40}{8} = 5 \) 2. \( x = \frac{-2}{8} = -\frac{1}{4} \) Therefore, the values of \( x \) are: \[ \boxed{5, -\frac{1}{4}}. \]