If | * 20 c 1&1&1 a&b&c a^2 & b^2 & c^2 | = 5 , then | * 20 c b c… - Vidyadip

MCQ

If $\left| {\begin{array}{*{20}{c}}
1&1&1\\
a&b&c\\
{{a^2}}&{{b^2}}&{{c^2}}
\end{array}} \right| = 5$ , then $\left| {\begin{array}{*{20}{c}}
{b{c^2} - {b^2}c}&{{a^2}c - a{c^2}}&{a{b^2} - b{a^2}}\\
{{b^2} - {c^2}}&{{c^2} - {a^2}}&{{a^2} - {b^2}}\\
{c - b}&{a - c}&{b - a}
\end{array}} \right|$ is equal to

A
$5$
B
$15$
✓
$25$
D
$35$

✓

Answer

Correct option: C.

$25$

c
| cofactor matrix of $A$ | $= | A |^2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 3 and 4 . determinant and metrices→3 and 4 . determinant and metrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2}, \frac{1}{6}$ and $\frac{1}{3}$, respectively. Each team gets $3$ points for a win, $1$ point for a draw and $0$ point for a loss in a game. Let $X$ and $Y$ denote the total points scored ky teams $T_1$ and $T_2$, respectively, after two games.

($1$) $P(X>Y)$ is

($A$) $\frac{1}{4}$ ($B$) $\frac{5}{12}$ ($C$) $\frac{1}{2}$ ($D$) $\frac{7}{12}$

($2$) $P(X=Y)$ is

($A$). $\frac{11}{36}$ ($B$) $\frac{1}{3}$ ($C$) $\frac{13}{36}$ ($D$) $\frac{1}{2}$

Given the answer quetion ($1$) and ($2$)

The value of $b$ for which the function $f(x)=\sin x-b x+c$ is strictly decreasing for $x \in R$ is given by

Let $f$ and $g$ be twice differentiable even functions on $(-2,2)$ such that $f\left(\frac{1}{4}\right)=0, f\left(\frac{1}{2}\right)=0, f(1)=1$ and $g\left(\frac{3}{4}\right)=0, g(1)=2$ Then, the minimum number of solutions of $f(x) g^{\prime \prime}(x)+f^{\prime}(x) g^{\prime}(x)=0$ in $(-2,2)$ is equal to

A disease affects two-thirds of the population of a country. A test for the disease gives the correct outcome with probability $\frac{2}{3}$. A person $X$ tested positive for the disease. The probability that $X$ has disease is

The general solution of the differential equation $\frac{{dy}}{{dx}} + \sin \left( {\frac{{x + y}}{2}} \right) = \sin \left( {\frac{{x - y}}{2}} \right)$ is

Let $f(x)=\int_0^x g(t) \log _e\left(\frac{1-t}{1+t}\right) d t$, where $g$ is a continuous odd function. If $\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$, then $\alpha$ is equal to..............

${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right) = $

The area under the curve y= 2x³+ 4x² between x = 2, x = 4 is:

192.6
198.6
88.3
172.3

$\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$ is equal to:

$\log\tan\Big(\frac{\pi}{3}+\frac{\pi}{2}\Big)+\text{C}$
$\log\tan\Big(\frac{\pi}{2}-\frac{\pi}{3}\Big)+\text{C}$
$\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$
None of these.

$I = \int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ is equal to