Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
For a real number $\alpha$, if the system $$\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$$ of linear equations, has infinitely many solutions, then $1 + \alpha + \alpha^2 = $
Step-by-Step Solution
Key Concept: When a determinant equals zero, the system is singular; checking which value of $a$ gives no solutions determines the valid constraint.
Expand the given determinant equation $D=0$ and factor to obtain $(1-a^2)(1+a+a^2-2a^2-a) = 0$, which simplifies to $(1-a^2)^2 = 0$, giving $a = \pm 1$. For $a=1$, the system of linear equations has no solutions. For $a=-1$, we have $1+a+a^2 = 1$.
Correct Answer: 1