Let A be a 3 × 3 real matrix such that A 1 1 0 = 1 1 0 ; A 1 0 1 = - 1 0 1 and A 0 0 1 = 1 1 2 . If X =…

Let A be a 3×3 real matrix such that A110=110;A101=-101 and A001=112. If X=x1   x2   x3T and I is an identity matrix of order 3, then the system A-2IX=411 has
  1. no solution
  2. infinitely many solutions
  3. unique solution
  4. exactly two solutions

Solution

Let the matrix be A=a1b1c1a2b2c2a3b3c3

Now A001=c1c2c3=112

  c1=1,c2=1,c3=2

A101=c1+a1c2+a2c3+a3=-101

  a1=-2,a2=-1,a3=-1

A110=a1+b1a2+b2a3+b3=110

  b1=3,b2=2,b3=1

So matrix A=-231-121-112

A-2I=-431-101-110

i.e. A-2 I=0

Now X=x1x2x3

So,   -431-101-110x1x2x3=411

-4x1+3x2+x3=4     ...1

-x1+x3=1       ...2

-x1+x2=1   ...3

Solving the above system of equations we get infinite solutions.

Asked in: JEE Main 2022 (25 Jun Shift 1)

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