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General Term of Binomial Expansion

Prove the fol...

Question

Prove the following identities:

$|\begin{matrix} 2 y & y - z - x & 2 y \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = {(x + y + z)}^{3}$

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Solution

$LHS = |\begin{matrix} 2 y & y - z - x & 2 y \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = |\begin{matrix} 2 y + 2 z + x - y - z & y - z - x + 2 z + 2 x & 2 y + z - x - y + 2 x \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = |\begin{matrix} x + y + z & x + y + z & x + y + z \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = (x + y + z) |\begin{matrix} 1 & 1 & 1 \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| [Taking (x + y + z) common from R_{1}] = (x + y + z) |\begin{matrix} 0 & 1 & 1 \\ 0 & 2 z & z - x - y \\ - x - y - z & 2 x & 2 x \end{matrix}| [Applying C_{1} \to C_{1} - C_{2}] = {(x + y + z)}^{2} |\begin{matrix} 0 & 1 & 1 \\ 0 & 2 z & z - x - y \\ - 1 & 2 x & 2 x \end{matrix}| [Taking (x + y + z) common from C_{1}] = {(x + y + z)}^{2} [- 1 (z - x - y - 2 z)] [Expanding along first column] = {(x + y + z)}^{3} = RHS ∴ |\begin{matrix} 2 y & y - z - x & 2 y \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = {(x + y + z)}^{3}$

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