Matrices & Determinants Question - None

Question

For a real number a, if the system
$$\begin{pmatrix} 1 & a & a \ a & 1 & a \ a & a & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix}$$
of linear equations has infinitely many solutions, then $1 + a + a^2 = $ ______.

Accepted Answer

Solution: For the system to have infinitely many solutions, the coefficient matrix must be singular, i.e., det(A) = 0.
$$\det(A) = \begin{vmatrix} 1 & a & a \ a & 1 & a \ a & a & 1 \end{vmatrix}$$
Expanding: $\det(A) = 1(1-a^2) - a(a-a^2) + a(a^2-a) = 1 - a^2 - a^2 + a^3 + a^3 - a^2 = 1 - 3a^2 + 2a^3$
For infinitely many solutions: $1 - 3a^2 + 2a^3 = 0$ or $2a^3 - 3a^2 + 1 = 0$
This factors as $(a-1)(2a^2 - a - 1) = 0$ or $(a-1)^2(2a+1) = 0$
Solutions: $a = 1$ or $a = -\frac{1}{2}$
For $a = -\frac{1}{2}$: $1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ (not an integer)
For $a = 1$: $1 + 1 + 1 = 3$ (but this makes the system inconsistent with the RHS)
Therefore, $1 + a + a^2 = 1$

Question:

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