Question

The system of equations \[ \begin{cases} x-2y+\alpha z=0,
2x+y-4z=0,
x-y+z=0 \end{cases} \] has a non-trivial solution for \(\alpha=\;\underline{\hspace{1cm}}\). \;(Answer in integer)

Hint:

"Non-trivial solution" for a homogeneous linear system means \(\det(A)=0\). Compute the determinant symbolically and solve for the parameter.

Answer 1

\[ \textbf{Step 1: Write the coefficient matrix and determinant.} \] \[ A=\begin{bmatrix} 1 & -2 & \alpha \\ 2 & 1 & -4 \\ 1 & -1 & 1 \end{bmatrix}, \quad \text{A non-trivial solution exists } \iff \det(A)=0. \] \[ \textbf{Step 2: Compute } \det(A). \] \[ \det(A) = \begin{vmatrix} 1 & -2 & \alpha \\ 2 & 1 & -4 \\ 1 & -1 & 1 \end{vmatrix} = 1\begin{vmatrix} 1 & -4 \\ -1 & 1 \end{vmatrix} - (-2)\begin{vmatrix} 2 & -4 \\ 1 & 1 \end{vmatrix} + \alpha \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}. \] \[ = 1(1-4) + 2(2+4) + \alpha(-2-1) = -3 + 12 - 3\alpha = 9 - 3\alpha. \] \[ \textbf{Step 3: Set determinant to zero.} \] \[ 9 - 3\alpha = 0 \;\;\Rightarrow\;\; \alpha = \boxed{3}. \]