Delta = left| {,begin{array}{*{20}{c}}{a + x}&b&cb&{x + c}&ac&a&{x + b}end{a

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

medium

$\Delta = \left| {\,\begin{array}{*{20}{c}}{a + x}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right|$,which of the following is a factor for the above determinant

A

$x - (a + b + c)$

B

$x + (a + b + c)$

C

$a + b + c$

D

$ - (a + b + c)$

Solution

(b) Operating ${C_1} \to {C_1} + {C_2} + {C_3}$
we find that $x + a + b + c$ is a factor.

Standard 12

Mathematics

Share

Similar Questions

Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ – 3i}&1\\4&{3i}&{ – 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then

easy

(IIT-1998)

$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ – 6}&{ – 2}\\
{ – 6}&{9 + {x^2}}&3\\
{ – 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by

normal

If the system of linear equations

$2 x+y-z=3$

$x-y-z=\alpha$

$3 x+3 y+\beta z=3$

has infinitely many solution, then $\alpha+\beta-\alpha \beta$ is equal to …. .

medium

(JEE MAIN-2021)

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}x&0&8\\4&1&3\\2&0&x\end{array}\,} \right| = 0$ are equal to

easy

The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third vertex which lies on line $y = x + 3$ is-

normal