Let 1, omega and omega^2 be the cube roots of | vedclass

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(A) The matrix is $M = \begin{bmatrix} 1 & a & b \ \omega & 1 & c \ \omega^2 & \omega & 1 \end{bmatrix}$.For the matrix to be non-singular,its deter

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$2$

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(A) The matrix is $M = \begin{bmatrix} 1 & a & b \ \omega & 1 & c \ \omega^2 & \omega & 1 \end{bmatrix}$.For the matrix to be non-singular,its determinant must be non-zero: $\det(M) 
eq 0$.Calculating the determinant: $\det(M) = 1(1 - \omega c) - a(\omega - \omega^2 c) + b(\omega^2 - \omega^2) = 1 - \omega c - a\omega + a\omega^2 c = 1 - \omega(a + c) + a\omega^2 c$.Given $a, c \in \{\omega, \omega^2\}$,we test the combinations:$1$. If $a = \omega, c = \omega$: $\det(M) = 1 - \omega(2\omega) + \omega(\omega^2)(\omega) = 1 - 2\omega^2 + \omega^4 = 1 - 2\omega^2 + \omega = 1 - 2\omega^2 + (-1 - \omega^2) = -3\omega^2 
eq 0$.$2$. If $a = \omega^2, c = \omega^2$: $\det(M) = 1 - \omega(2\omega^2) + \omega^2(\omega^2)(\omega^2) = 1 - 2\omega^3 + \omega^6 = 1 - 2(1) + 1 = 0$.$3$. If $a = \omega, c = \omega^2$: $\det(M) = 1 - \omega(\omega + \omega^2) + \omega(\omega^2)(\omega^2) = 1 - \omega(-1) + \omega^5 = 1 + \omega + \omega^2 = 0$.$4$. If $a = \omega^2, c = \omega$: $\det(M) = 1 - \omega(\omega^2 + \omega) + \omega^2(\omega^2)(\omega) = 1 - \omega(-1) + \omega^5 = 1 + \omega + \omega^2 = 0$.Since $b$ can be either $\omega$ or $\omega^2$ in the case where $\det(M) 
eq 0$,and only the case $(a, c) = (\omega, \omega)$ yields a non-zero determinant,there are $2$ possible values for $b$ (namely $\omega$ and $\omega^2$).Thus,the number of elements in $S$ is $2$.

Let $1, \omega$ and $\omega^2$ be the cube roots of unity. If $S$ is the set of all non-singular matrices of the form $M = \begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}$ where $a, b, c \in \{\omega, \omega^2\}$,then the number of elements in $S$ is

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