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

Vedclass · Accepted Answer

$4$

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 $a$ common from $C_1$,$b$ common from $C_2$,and $c$ common from $C_3$:
$\Delta = (abc)(abc) \left|\begin{array}{ccc}0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0\end{array}ight| = (abc)^2 \left|\begin{array}{ccc}0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0\end{array}ight|$.
Expanding the determinant:
$\Delta = (abc)^2 [0(0-1) - 1(0-1) + 1(1-0)] = (abc)^2 [0 + 1 + 1] = 2(abc)^2$.
Comparing with $m(abc)^k$,we get $m = 2$ and $k = 2$.
Therefore,$m+k = 2+2 = 4$.

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

$\left|\begin{array}{cc}\sin \frac{2 \pi}{9} & \cos \frac{2 \pi}{9} \\ \sin \frac{5 \pi}{18} & \cos \frac{5 \pi}{18}\end{array}\right|=$ . . . . . . .

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, then $\det(A)$ is equal to

If $\left|\begin{array}{ccc}x & 4 & 6 \\ 2 & 3 & -9 \\ 5 & 6 & 1\end{array}\right|+\left|\begin{array}{ccc}5 & 6 & 1 \\ 6 & 4 & 5 \\ 2 & 3 & -9\end{array}\right|=\left|\begin{array}{ccc}2 & 3 & -9 \\ 1-2 x & -8 & -11 \\ 5 & 6 & 1\end{array}\right|$,then $x=$ . . . . . .

The area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

If $\alpha, \beta, \gamma$ are the roots of $\left|\begin{array}{ccc} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0$,then $\alpha \beta+\beta \gamma+\gamma \alpha=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module