If $\left| {\begin{array}{*{20}{c}} {x + 1}&{x + 2}&{x + 3}\\ {x + 2}&{x + 3}&{x + 4}\\ {x + a}&{x + b}&{x + c} \end{array}} \right| = 0$ then $a,b,c$ are in

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Analyzing +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} {x + 1}&{x + 2}&{x + 3}\\ {x + 2}&{x + 3}&{x + 4}\\ {x + a}&{x + b}&{x + c} \end{array}} \right| = 0

then

a,b,c

are in

View answer and explanation

Correct option: A

Option A (AP) is correct because after simplifying the determinant, we find that 2b = a + c, which is the condition for a, b, c to be in Arithmetic Progression.

More Questions from MATRICES

Practice All Chapter MCQs →

\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

and

a, b, c

are real numbers then the line

a x+b y+c=0

passes through the fixed point is The number of solutions of the system of equations

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

is If the equations

2x + 3y + 1 = 0,3x + y - 2 = 0

and

ax + 2y - b = 0

are consistent, then The value of the determinant

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

, is If

\omega

is the cube root of unity, then

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

is equal to For any matrix

A

of order

2 \times 2

, if

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

, then

|A|

is equal to