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(D) Given the determinant: $\left| \begin{matrix} 1 & a & a^2 \ 1 & x & x^2 \ b^2 & ab & a^2 \end{matrix} ight| = 0$. Applying row operations $R_2

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both $(A)$ and $(C)$

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(D) Given the determinant: $\left| \begin{matrix} 1 & a & a^2 \ 1 & x & x^2 \ b^2 & ab & a^2 \end{matrix} ight| = 0$. Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $\left| \begin{matrix} 1 & a & a^2 \ 0 & x-a & x^2-a^2 \ b^2-1 & ab-a & 0 \end{matrix} ight| = 0$. Factoring out $(x-a)$ from $R_2$ and $(b-1)$ from $R_3$: $(x-a)(b-1) \left| \begin{matrix} 1 & a & a^2 \ 0 & 1 & x+a \ b+1 & a & 0 \end{matrix} ight| = 0$. Expanding along $C_1$: $(x-a)(b-1) [1(0 - a(x+a)) - 0 + (b+1)(a(x+a) - a^2)] = 0$. $(x-a)(b-1) [-ax - a^2 + (b+1)(ax)] = 0$. $(x-a)(b-1) [-ax - a^2 + abx + ax] = 0$. $(x-a)(b-1) [abx - a^2] = 0$. $(x-a)(b-1) a(bx - a) = 0$. This gives $x = a$ or $x = \frac{a}{b}$.

If $\left| \begin{matrix} 1 & a & a^2 \\ 1 & x & x^2 \\ b^2 & ab & a^2 \end{matrix} \right| = 0$,then:

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