left| {begin{array}{ccc} 1/a & a^2 & bc 1/b | vedclass

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

Question

(D) To evaluate the determinant $\Delta = \left| {\begin{array}{ccc} 1/a & a^2 & bc \ 1/b & b^2 & ca \ 1/c & c^2 & ab \end{array}} ight|$,we multi

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) To evaluate the determinant $\Delta = \left| {\begin{array}{ccc} 1/a & a^2 & bc \ 1/b & b^2 & ca \ 1/c & c^2 & ab \end{array}} ight|$,we multiply and divide by $abc$:$\Delta = \frac{1}{abc} \left| {\begin{array}{ccc} 1 & a^3 & abc \ 1 & b^3 & abc \ 1 & c^3 & abc \end{array}} ight|$Now,take $abc$ common from the third column:$\Delta = \frac{abc}{abc} \left| {\begin{array}{ccc} 1 & a^3 & 1 \ 1 & b^3 & 1 \ 1 & c^3 & 1 \end{array}} ight|$Since the first column and the third column are identical,the value of the determinant is $0$.

$\left| {\begin{array}{ccc} 1/a & a^2 & bc \\ 1/b & b^2 & ca \\ 1/c & c^2 & ab \end{array}} \right| = $

Similar Questions

If $\begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix} = (x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then find the value of $k$.

If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =

If $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ a & b & c \\ l & m & n\end{array}\right|=(-1)^K\left|\begin{array}{ccc}m & n & l \\ b & c & a \\ \beta & \gamma & \alpha\end{array}\right|$,then the least value of $K$ is

The value of $\left|\begin{array}{lll}1990 & 1991 & 1992 \\ 1991 & 1992 & 1993 \\ 1992 & 1993 & 1994\end{array}\right|$ is

The determinant of a skew-symmetric matrix of order $3$ is always:

Vedclass Test Series

Exam Paper Generator

Online Exam Module