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Factorization

If 2-x 2+x 2+...

Question

If $|\begin{matrix} 2 - x & 2 + x & 2 + x \\ 2 + x & 2 - x & 2 + x \\ 2 + x & 2 + x & 2 - x \end{matrix}| = 0,$ then x = ________________.

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Solution

Given: $|\begin{matrix} 2 - x & 2 + x & 2 + x \\ 2 + x & 2 - x & 2 + x \\ 2 + x & 2 + x & 2 - x \end{matrix}| = 0$

$|\begin{matrix} 2 - x & 2 + x & 2 + x \\ 2 + x & 2 - x & 2 + x \\ 2 + x & 2 + x & 2 - x \end{matrix}| = 0 Applying R_{2} \to R_{2} - R_{1} \Rightarrow |\begin{matrix} 2 - x & 2 + x & 2 + x \\ 2 + x - 2 + x & 2 - x - 2 - x & 2 + x - 2 - x \\ 2 + x & 2 + x & 2 - x \end{matrix}| = 0 \Rightarrow |\begin{matrix} 2 - x & 2 + x & 2 + x \\ 2 x & - 2 x & 0 \\ 2 + x & 2 + x & 2 - x \end{matrix}| = 0 Taking (2 x) common from R_{2} \Rightarrow (2 x) |\begin{matrix} 2 - x & 2 + x & 2 + x \\ 1 & - 1 & 0 \\ 2 + x & 2 + x & 2 - x \end{matrix}| = 0 Applying R_{3} \to R_{3} - R_{1} \Rightarrow (2 x) |\begin{matrix} 2 - x & 2 + x & 2 + x \\ 1 & - 1 & 0 \\ 2 + x - 2 + x & 2 + x - 2 - x & 2 - x - 2 - x \end{matrix}| = 0 \Rightarrow (2 x) |\begin{matrix} 2 - x & 2 + x & 2 + x \\ 1 & - 1 & 0 \\ 2 x & 0 & - 2 x \end{matrix}| = 0 Taking (2 x) common from R_{3} \Rightarrow {(2 x)}^{2} |\begin{matrix} 2 - x & 2 + x & 2 + x \\ 1 & - 1 & 0 \\ 1 & 0 & - 1 \end{matrix}| = 0 Applying C_{3} \to C_{3} + C_{1} \Rightarrow {(2 x)}^{2} |\begin{matrix} 2 - x & 2 + x & 2 + x + 2 - x \\ 1 & - 1 & 0 + 1 \\ 1 & 0 & - 1 + 1 \end{matrix}| = 0 \Rightarrow {(2 x)}^{2} |\begin{matrix} 2 - x & 2 + x & 4 \\ 1 & - 1 & 1 \\ 1 & 0 & 0 \end{matrix}| = 0 Expanding through R_{3} \Rightarrow {(2 x)}^{2} [(1) ((2 + x) (1) + 4)] = 0 \Rightarrow {(2 x)}^{2} [(2 + x + 4)] = 0 \Rightarrow {(2 x)}^{2} (x + 6) = 0 \Rightarrow {(2 x)}^{2} = 0 or (6 + x) = 0 \Rightarrow x = 0 or x = - 6$

Hence, x = 0, −6.

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