If left|begin{array}{ccc}0 & ab^2 & ac^2 a^2 | vedclass

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

Question

(D) Let $\Delta = \left|\begin{array}{ccc}0 & ab^2 & ac^2 \ a^2b & 0 & bc^2 \ a^2c & b^2c & 0\end{array}ight|$.Taking $a$ common from $R_1$, $b$ c

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let $\Delta = \left|\begin{array}{ccc}0 & ab^2 & ac^2 \ a^2b & 0 & bc^2 \ a^2c & b^2c & 0\end{array}ight|$.Taking $a$ common from $R_1$, $b$ common from $R_2$, and $c$ common from $R_3$:$\Delta = abc \left|\begin{array}{ccc}0 & b^2 & c^2 \ a^2 & 0 & c^2 \ a^2 & b^2 & 0\end{array}ight|$.Now, taking $a^2$ common from $C_1$, $b^2$ common from $C_2$, and $c^2$ common from $C_3$:$\Delta = abc(a^2 b^2 c^2) \left|\begin{array}{ccc}0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0\end{array}ight| = (abc)^3 \left|\begin{array}{ccc}0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0\end{array}ight|$.Expanding the determinant:$\Delta = (abc)^3 [0(0-1) - 1(0-1) + 1(1-0)] = (abc)^3 [0 + 1 + 1] = 2(abc)^3$.Comparing with $m(abc)^k$, we get $m = 2$ and $k = 3$.Therefore, $m+k = 2+3 = 5$.

If $\left|\begin{array}{ccc}0 & ab^2 & ac^2 \\ a^2b & 0 & bc^2 \\ a^2c & b^2c & 0\end{array}\right|=m(abc)^k$, then $m+k=$ . . . . . . .

Similar Questions

If ${\Delta _1} = \left| {\begin{array}{{20}{c}}1&0\\a&b\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{{20}{c}}1&0\\c&d\end{array}} \right|$,then ${\Delta _2}{\Delta _1}$ is equal to

The area of a triangle with vertices $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$ is $6$ sq. units. Find the value of $k$.

The set of all values of $t \in R$,for which the matrix $\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]$ is invertible.

For positive numbers $x, y$ and $z$,the numerical value of the determinant $\left| \begin{array}{ccc} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{array} \right|$ is

If $ax^4+bx^3+cx^2+50x+d = \begin{vmatrix} x^3-14x^2 & -x & 3x+\lambda \\ 4x+1 & 3x & x-4 \\ -3 & 4 & 0 \end{vmatrix}$,then find $\lambda$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module