Let omega neq 1 be a cube root of unity and S | vedclass

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Question

(A) Let $A = \begin{bmatrix} 1 & a & b \ \omega & 1 & c \ \omega^2 & \omega & 1 \end{bmatrix}$. For the matrix to be non-singular,$|A| 
eq 0$.$|A|

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $A = \begin{bmatrix} 1 & a & b \ \omega & 1 & c \ \omega^2 & \omega & 1 \end{bmatrix}$. For the matrix to be non-singular,$|A| 
eq 0$.$|A| = 1(1 - \omega c) - a(\omega - \omega^2 c) + b(\omega^2 - \omega^2) = 1 - \omega c - a\omega + a\omega^2 c = (1 - \omega c) - a\omega(1 - \omega c) = (1 - \omega c)(1 - a\omega)$.For $|A| 
eq 0$,we must have $(1 - \omega c) 
eq 0$ and $(1 - a\omega) 
eq 0$.This implies $c 
eq \frac{1}{\omega} = \omega^2$ and $a 
eq \frac{1}{\omega} = \omega^2$.Since $a, b, c \in \{\omega, \omega^2\}$,the condition $a 
eq \omega^2$ implies $a = \omega$,and $c 
eq \omega^2$ implies $c = \omega$.The variable $b$ can be either $\omega$ or $\omega^2$.Thus,the possible matrices are $\begin{bmatrix} 1 & \omega & \omega \ \omega & 1 & \omega \ \omega^2 & \omega & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & \omega & \omega^2 \ \omega & 1 & \omega \ \omega^2 & \omega & 1 \end{bmatrix}$.Therefore,the number of distinct matrices in the set $S$ is $2$.

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