The determinant left| begin{array}{ccc} ^x{C_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $^n{C_r} = \frac{n(n-1)...(n-r+1)}{r!}$.Substituting these values into the determinant:$\Delta = \left| \begin{array}{ccc} x & \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that $^n{C_r} = \frac{n(n-1)...(n-r+1)}{r!}$.Substituting these values into the determinant:$\Delta = \left| \begin{array}{ccc} x & \frac{x(x-1)}{2} & \frac{x(x-1)(x-2)}{6} \ y & \frac{y(y-1)}{2} & \frac{y(y-1)(y-2)}{6} \ z & \frac{z(z-1)}{2} & \frac{z(z-1)(z-2)}{6} \end{array} ight|$Taking $\frac{x}{1}, \frac{y}{2}, \frac{z}{6}$ common from the columns respectively:$\Delta = \frac{xyz}{12} \left| \begin{array}{ccc} 1 & x-1 & (x-1)(x-2) \ 1 & y-1 & (y-1)(y-2) \ 1 & z-1 & (z-1)(z-2) \end{array} ight|$Applying $C_2 ightarrow C_2 + C_1$:$\Delta = \frac{xyz}{12} \left| \begin{array}{ccc} 1 & x & x^2-3x+2 \ 1 & y & y^2-3y+2 \ 1 & z & z^2-3z+2 \end{array} ight|$Using column operations $C_3 ightarrow C_3 + 3C_2 - 2C_1$,we reduce the determinant to the standard Vandermonde form:$\Delta = \frac{xyz}{12} \left| \begin{array}{ccc} 1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2 \end{array} ight| = \frac{xyz}{12} (x-y)(y-z)(z-x)$.

The determinant $\left| \begin{array}{ccc} ^x{C_1} & ^x{C_2} & ^x{C_3} \\ ^y{C_1} & ^y{C_2} & ^y{C_3} \\ ^z{C_1} & ^z{C_2} & ^z{C_3} \end{array} \right|$ equals:

Similar Questions

If $f(x) = \left| \begin{array}{ccc} 2\cos^2 2x & \sin 2x & -\sin x \\ \sin 2x & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array} \right|$,then the value of $\int_{0}^{\frac{\pi}{2}} f'(x) \,dx$ is equal to

The value of the determinant $\left| \begin{array}{ccc} a^2 & a & 1 \\ \cos(nx) & \cos(n+1)x & \cos(n+2)x \\ \sin(nx) & \sin(n+1)x & \sin(n+2)x \end{array} \right|$ is independent of :

Let $f(x) = \left| \begin{array}{ccc} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$,where $p$ is a constant. Then $\frac{d^3}{dx^3} \{f(x)\}$ at $x = 0$ is

If $A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$ and the rank of $A$ is $2$,then the value of $x$ is equal to

If the matrix $A = \begin{bmatrix} 1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & \alpha \end{bmatrix}$ is of rank $3$, then $\alpha$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module