Home
Class 12
MATHS
If A=[{:(8,-6,2),(-6,7,-4),(2,-4,3):}] a...

If `A=[{:(8,-6,2),(-6,7,-4),(2,-4,3):}]` and `X` is a non zero column matrix such that `AX=lambdaX`, where `lambda` is a scalar, then values of `lambda` can be

A

`3`

B

`6`

C

`12`

D

`15`

Text Solution

Verified by Experts

The correct Answer is:
A, D

`(a,d)` Let `X=[{:(a),(b),(c):}]`
`:.AX=lambdaXimplies[{:(8,-6,2),(-6,7,-4),(2,-4,3):}][{:(a),(b),(c ):}]=[{:(lambdaa),(lambdab),(lambdac ):}]`
`implies(8-lambda)a-6b+2c=0`
`-6a+(7-lambda)b-4c=0`
and `2a-4b+(3-lambda)c=0`
For non zero solution `[{:(8-lambda,,-6,,2),(-6,,7-lambda,,-4),(2,,-4,,3-lambda):}]=0`
`implieslambda=0,3,15`
Promotional Banner

Topper's Solved these Questions

  • MATRICES

    CENGAGE|Exercise Single correct Answer|34 Videos
  • MATHMETICAL REASONING

    CENGAGE|Exercise JEE Previous Year|10 Videos
  • METHODS OF DIFFERENTIATION

    CENGAGE|Exercise Multiple Correct Answer Type|7 Videos

Similar Questions

Explore conceptually related problems

If X is a non-zero column matrix, such that A X=lambda X where lambda is a scalar and the matrix A is [[4,6,6],[1,3,2],[-1,-5,-2]] then sum of distinct values of lambda is

If A=[{:(8,-6,2),(-6,7,-4),(2,-4,3):}] verify that A(adjA)=(adjA)A=A|A|I_(3) .

If |vec a| = 4 and -3 le lambda le 2 then the range of |lambda vec a|

If A=[(3,-2),(4,-1)] , then find all the possible values of lambda such that the matrix (A-lambdaI) is singular.

If (1,2,4) and (2,-3 lambda,-3 ) are the initial and terminal points of the vector hati+5hatj-7hatk , then value of lambda is equal to

The matrix [(5,10,3),(-2,-4,6),(-1,-2,x)] is a singular matrix if the value of x is

If f(x)=lambda|sinx|+lambda^2|cosx|+g(lambda) has a period = pi/2 then find the value of lambda

If (3)/(2)+(7)/(2)i is a solution of the equation ax^(2)-6x+b=0 , where a and b are real numbers, then the value of a+b is equal to

If the equation |x^2-5x + 6|-lambda x+7 lambda=0 has exactly 3 distinct solutions then lambda is equal to

An infinite number of tangents can be drawn from (1,2) to the circle x^2+y^2-2x-4y+lambda=0 . Then find the value of lambda