Aspire Faculty ID #13732 · Topic: JAMIA MILLIA ISLAMIA MCA 2022 · Just now
JAMIA MILLIA ISLAMIA MCA 2022

If $ \begin{vmatrix} x & 3 & 6 \\ 3 & 6 & x \\ 6 & x & 3 \end{vmatrix} = \begin{vmatrix} 2 & x & 7 \\ x & 7 & 2 \\ 7 & 2 & x \end{vmatrix} = \begin{vmatrix} 4 & 5 & x \\ 5 & x & 4 \\ x & 4 & 5 \end{vmatrix} = 0 $, then $x$ is equal to:

Solution

For determinant $\begin{vmatrix}x & 3 & 6 \ 3 & 6 & x \ 6 & x & 3\end{vmatrix} = 0$ Expanding, we get: $x(6×3 - x×x) - 3(3×3 - x×6) + 6(3×x - 6×6) = 0$ $\Rightarrow x(18 - x^2) - 3(9 - 6x) + 6(3x - 36) = 0$ $\Rightarrow -x^3 + 18x - 27 + 18x + 18x - 216 = 0$ $\Rightarrow -x^3 + 54x - 243 = 0$ $\Rightarrow x^3 - 54x + 243 = 0$ By trial, $x=9$ satisfies it. Hence $\boxed{x = 9}$

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