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(B) For the system of linear equations to have non-trivial solutions,the determinant of the coefficient matrix must be zero.$\Delta = \left|\begin{arr

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(B) For the system of linear equations to have non-trivial solutions,the determinant of the coefficient matrix must be zero.$\Delta = \left|\begin{array}{ccc} a & -a^2 & a^3 \ -a^2 & a^3 & a \ a^3 & a & -a^2 \end{array}ight| = 0$Taking $a$ common from each row:$\Delta = a^3 \left|\begin{array}{ccc} 1 & -a & a^2 \ -a & a^2 & 1 \ a^2 & 1 & -a \end{array}ight| = 0$Expanding the determinant:$a^3 [1(-a^3 - 1) + a(a^2 - a^2) + a^2(-a - a^4)] = 0$$a^3 [-a^3 - 1 - a^3 - a^6] = 0$$-a^3 (a^6 + 2a^3 + 1) = 0$$-a^3 (a^3 + 1)^2 = 0$Since $a^3 + 1 = (a+1)(a^2 - a + 1) = 0$,the roots are $a = -1$ and $a = -\omega, -\omega^2$,where $\omega$ is the cube root of unity.Given $a$ is a non-real complex number,$a = -\omega$ or $a = -\omega^2$.Since $|-\omega| = |-\omega^2| = 1$,we have $|a| = 1$.

If $a$ is a non-real complex number for which the system of equations $ax - a^2y + a^3z = 0$,$-a^2x + a^3y + az = 0$,and $a^3x + ay - a^2z = 0$ has non-trivial solutions,then $|a|$ is:

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