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Mathematics Question on Determinants

The value of ∣x+yy+zz+x\[0.3em]xyz\[0.3em]x−yy−zz−x∣\begin{vmatrix} x+y&y+z & z+ x \\\[0.3em] x & y & z \\\[0.3em] x-y & y-z & z-x \end{vmatrix} = is equal to :

A

00

B

(x+y+z)3(x+y+z)^3

C

2(x+y+z)32(x+y+z)^3

D

2(x+y+z)22 (x+y+z)^2

Answer

00

Explanation

Solution

Let A=[x+yy+zz+x xyz x−yy−zz−x]A = \begin{bmatrix}x+y&y+z&z+x\\\ x&y&z\\\ x-y&y-z&z-x\end{bmatrix}
Applying C1→C1+C2+C3C_{1} \to C_{1} + C_{2} +C_{3}
=[2(x+y+z)y+zz+x x+y+zyz 0y−zz−x]= \begin{bmatrix}2\left(x+y+z\right)&y+z&z+x\\\ x+y+z&y&z\\\ 0&y-z&z-x\end{bmatrix}
=(x+y+z)[2y+zz+x 1yz 0y−zz−x]= \left(x+y+z\right) \begin{bmatrix}2&y+z&z+x\\\ 1&y&z\\\ 0&y-z&z-x\end{bmatrix}
Applying R2→2R2−R1R_{2} \to 2R_{2} -R_{1}
=(x+y+z)[2y+zz+x 0y−zz−x 0y−zz−x]= \left(x+y+z\right) \begin{bmatrix}2&y+z&z+x\\\ 0&y-z&z-x\\\ 0&y-z&z-x\end{bmatrix}
=0= 0 (∵\because Two rows are identical)