If left| {begin{array}{*{20}{c}}{ - {a^2}}&{a | vedclass

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

Question

(C) Given determinant is $\Delta = \left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\{ab}&{ - {b^2}}&{bc}\{ac}&{bc}&{ - {c^2}}\end{array}} ight|

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given determinant is $\Delta = \left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\{ab}&{ - {b^2}}&{bc}\{ac}&{bc}&{ - {c^2}}\end{array}} ight|$.Taking $a, b, c$ common from $R_1, R_2, R_3$ respectively:$\Delta = (abc) \left| {\begin{array}{*{20}{c}}{ - a}&b&c\a&{ - b}&c\a&b&{ - c}\end{array}} ight|$.Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:$\Delta = (abc)(abc) \left| {\begin{array}{*{20}{c}}{ - 1}&1&1\1&{ - 1}&1\1&1&{ - 1}\end{array}} ight| = {a^2}{b^2}{c^2} \left| {\begin{array}{*{20}{c}}{ - 1}&1&1\1&{ - 1}&1\1&1&{ - 1}\end{array}} ight|$.Now,expanding the determinant:$\left| {\begin{array}{*{20}{c}}{ - 1}&1&1\1&{ - 1}&1\1&1&{ - 1}\end{array}} ight| = -1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) = 0 + 2 + 2 = 4$.Therefore,$\Delta = 4{a^2}{b^2}{c^2}$.Comparing with $K{a^2}{b^2}{c^2}$,we get $K = 4$.

If $\left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}} \right| = K{a^2}{b^2}{c^2},$ then $K = $

Similar Questions

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

If $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \\ 0 & x-3 & 3x^2-27 \\ 0 & 2x-4 & 8x^2-32 \end{array} \right|$,then $\lim_{x \rightarrow 1} \frac{f(x)}{f(-x)} = $

The solution$(s)$ of the equation $\left| \begin{matrix} x & a & b \\ a & x & a \\ b & b & x \end{matrix} \right| = 0$ is/are:

If $t_1, t_2$ and $t_3$ are distinct,then the points $(t_1, 2at_1 + at_1^3)$,$(t_2, 2at_2 + at_2^3)$ and $(t_3, 2at_3 + at_3^3)$ are collinear if:

If $\left|\begin{array}{ccc}9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36\end{array}\right|=K$,then $K, K+1$ are the roots of the equation

Vedclass Test Series

Exam Paper Generator

Online Exam Module