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Question: The determinant $\begin{vmatrix} ^{x}C_1 & ^{x}C_2 & ^{x}C_3 \\ ^{y}C_1 & ^{y}C_2 & ^{y}C_3 \\ ^{z}C...

The determinant ∣xC1xC2xC3yC1yC2yC3zC1zC2zC3∣\begin{vmatrix} ^{x}C_1 & ^{x}C_2 & ^{x}C_3 \\ ^{y}C_1 & ^{y}C_2 & ^{y}C_3 \\ ^{z}C_1 & ^{z}C_2 & ^{z}C_3 \end{vmatrix}

A

13\frac{1}{3}xyz(x+y)(y+z)(z+x)

B

14\frac{1}{4}xyz(x+y-z)(y+z-x)

C

112\frac{1}{12}xyz(x-y)(y-z)(z-x)

D

none

Answer

112xyz(x−y)(y−z)(z−x)\frac{1}{12}xyz(x-y)(y-z)(z-x)

Explanation

Solution

The given determinant is

∣xC1xC2xC3yC1yC2yC3zC1zC2zC3∣\begin{vmatrix} {^xC_1} & {^xC_2} & {^xC_3} \\ {^yC_1} & {^yC_2} & {^yC_3} \\ {^zC_1} & {^zC_2} & {^zC_3} \end{vmatrix}

Recall that

nC1=n,nC2=n(n−1)2,nC3=n(n−1)(n−2)6.{^nC_1}=n,\quad {^nC_2}=\frac{n(n-1)}{2},\quad {^nC_3}=\frac{n(n-1)(n-2)}{6}.

Thus, the matrix becomes

(xx(x−1)2x(x−1)(x−2)6yy(y−1)2y(y−1)(y−2)6zz(z−1)2z(z−1)(z−2)6).\begin{pmatrix} x & \dfrac{x(x-1)}{2} & \dfrac{x(x-1)(x-2)}{6}\\[1mm] y & \dfrac{y(y-1)}{2} & \dfrac{y(y-1)(y-2)}{6}\\[1mm] z & \dfrac{z(z-1)}{2} & \dfrac{z(z-1)(z-2)}{6} \end{pmatrix}.

Step 1. Factor out xx, yy, zz from rows 1, 2, 3 respectively:

=x y z⋅∣1x−12(x−1)(x−2)61y−12(y−1)(y−2)61z−12(z−1)(z−2)6∣.= x\,y\,z \cdot \begin{vmatrix} 1 & \dfrac{x-1}{2} & \dfrac{(x-1)(x-2)}{6}\\[1mm] 1 & \dfrac{y-1}{2} & \dfrac{(y-1)(y-2)}{6}\\[1mm] 1 & \dfrac{z-1}{2} & \dfrac{(z-1)(z-2)}{6} \end{vmatrix}.

Step 2. Factor out constants from columns 2 and 3: factor 12\frac{1}{2} from column 2 and 16\frac{1}{6} from column 3. The overall factor becomes 12⋅16=112\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}:

=x y z12⋅∣1x−1(x−1)(x−2)1y−1(y−1)(y−2)1z−1(z−1)(z−2)∣.= \frac{x\,y\,z}{12}\cdot \begin{vmatrix} 1 & x-1 & (x-1)(x-2)\\[1mm] 1 & y-1 & (y-1)(y-2)\\[1mm] 1 & z-1 & (z-1)(z-2) \end{vmatrix}.

Step 3. Substitute u=x−1,  v=y−1,  w=z−1u=x-1,\; v=y-1,\; w=z-1. Then the determinant becomes

D=x y z12⋅∣1uu(u+1)1vv(v+1)1ww(w+1)∣.D = \frac{x\,y\,z}{12}\cdot \begin{vmatrix} 1 & u & u(u+1)\\[1mm] 1 & v & v(v+1)\\[1mm] 1 & w & w(w+1) \end{vmatrix}.

Subtract the first row from the others:

R2→R2−R1,R3→R3−R1,R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_1,

yielding

D=x y z12⋅∣1uu(u+1)0v−uv(v+1)−u(u+1)0w−uw(w+1)−u(u+1)∣.D = \frac{x\,y\,z}{12}\cdot \begin{vmatrix} 1 & u & u(u+1)\\[1mm] 0 & v-u & v(v+1)-u(u+1)\\[1mm] 0 & w-u & w(w+1)-u(u+1) \end{vmatrix}.

Note that

v(v+1)−u(u+1)=(v2+v)−(u2+u)=(v−u)(v+u+1),v(v+1)-u(u+1) = (v^2+v)-(u^2+u) = (v-u)(v+u+1), w(w+1)−u(u+1)=(w−u)(w+u+1).w(w+1)-u(u+1) = (w-u)(w+u+1).

Thus, the 2×2 determinant becomes

(v−u)(w−u)[(w+u+1)−(v+u+1)]=(v−u)(w−u)(w−v).(v-u)(w-u)\left[(w+u+1)-(v+u+1)\right] = (v-u)(w-u)(w-v).

Returning to original variables:

v−u=(y−1)−(x−1)=y−x,w−u=z−x,w−v=z−y.v-u = (y-1) - (x-1)= y-x,\quad w-u = z-x,\quad w-v = z-y.

So,

D=x y z12 (y−x)(z−x)(z−y).D = \frac{x\,y\,z}{12}\,(y-x)(z-x)(z-y).

Observe that

(y−x)(z−x)(z−y)=(x−y)(y−z)(z−x)(y-x)(z-x)(z-y) = (x-y)(y-z)(z-x)

(up to rearrangement of factors, which does not change the product since the number of sign changes is even).

Final Answer:

112 x y z (x−y)(y−z)(z−x)\boxed{\frac{1}{12}\,x\,y\,z\,(x-y)(y-z)(z-x)}