If | * 20 c a - b - c & 2a & 2a 2b & b - c - a & 2b 2c & 2c & c -… - Vidyadip

MCQ

If $\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to

A
$abc$
✓
$ - 2 \left( {a + b + c} \right)$
C
$ 2 \left( {a + b + c} \right)$
D
$ - \left( {a + b + c} \right)$

✓

Answer

Correct option: B.

$ - 2 \left( {a + b + c} \right)$

b
${R_1} \to {R_1} + {R_2} + {R_3}$

$\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&1&1\\
{2b}&{b - a - c}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$

${c_3} \to {c_3} - {c_1},{c_2} \to {c_2} - {c_1}$

$ = \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&0&0\\
{2b}&{ - \left( {a + b + c} \right)}&0\\
{2c}&0&{ - \left( {a + b + c} \right)}
\end{array}} \right|$

$ = {\left( {a + b + c} \right)^3}$

$ = \left( {a + b + c} \right){\left( {x + a + b + c} \right)^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

If $\vec{b}$ and $\vec{c}$ are any two non-collinear unit vectors and $\vec{a}$ is any vector, then find the value of $(\vec{a} \cdot \vec{b}) \vec{b}+(\vec{a} \cdot \vec{c}) \vec{c}+\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} \cdot(\vec{b} \times \vec{c})$.

If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $

The position vectors of the points A, B, C are $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ respectively. These points,

Form an isosceles triangle.
Form a right triangle.
Are collinear.
Form a scalene triangle.

If $\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4},$ then x is:

$\frac{1}{6}$
$1$
$(\frac{1}{6},-1)$
None of these.

$\int_{}^{} {{x^5}.{e^{{x^2}}}dx = } $

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 probability that a man can hit a target is $\frac{3}{4}$. He tries $5$ times. The probability that he will hit the target at least three times is

$\int {\frac{{{{\sin }^{ - 1}}x - {{\cos }^{ - 1}}x}}{{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x}}} dx = $

Let $\alpha$ (a) and $\beta$ (a) be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a>-1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

If ${a_1},{a_2},{a_3},........,{a_n},......$ are in G.P. and ${a_i} > 0$ for each $i$, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right|$ is equal to