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Mathematics Question on Determinants

If ∣y+zx−zx−y y−zz+xy−x z−yz−xx+y∣=kxyz\begin{vmatrix}y+z&x-z&x-y\\\ y-z &z+x&y-x\\\ z-y &z-x&x+y\end{vmatrix}= kxyz then the value of k is :

A

2

B

4

C

6

D

8

Answer

8

Explanation

Solution

Let ∣y+zx−zx−y y−zz+xy−x z−yz−xx+y∣\begin{vmatrix}y+z&x-z&x-y\\\ y-z &z+x&y-x\\\ z-y &z-x&x+y\end{vmatrix} Applying C1→C1+C2+C3C_{1} \to C_{1} + C_{2} +C_{3} =∣2xx−zx−y 2yz+xy−x 2zz−xx+y∣ = \begin{vmatrix}2x&x-z&x-y\\\ 2y &z+x&y-x\\\ 2z &z-x&x+y\end{vmatrix} Applying R1→R2+R1,R3→R3+R1 R_{1} \to R_{2}+R_{1}, R_{3} \to R_{3} + R_{1} =∣2xx−zx−y 2(x+y)2x0 2(z+z)02x∣= \begin{vmatrix}2x&x-z&x-y\\\ 2\left(x+y\right) &2x&0\\\ 2\left(z+z\right)&0&2x\end{vmatrix} On expanding we get =2x(4x2)−(x−z)[4x(x+y)]+(x−y)[−4x(x+z)]= 2x (4x^2) - (x - z) [4x (x + y)] + (x - y) [- 4x (x + z)] =8x2−(x−z)(4x2+4xy)−(x−y)(4x2+4xz)= 8x^2 - (x - z) (4x^2 + 4xy) - (x - y) (4x^2 + 4xz) =8x3−4x3−4x2y+4zx2+4xyz−4x3−4x2z+4yx2+4xyz= 8x^3 - 4x^3 - 4x^2y + 4^zx2 + 4xyz - 4x^3 - 4x^2z + 4yx^2 + 4xyz =8xyz= 8xyz Given : A=kxyz ⇒ 8xyz=kxyz ⇒ k=8 A = kxyz \ \Rightarrow \ 8xyz = kxyz \ \Rightarrow \ k = 8