Question
Let M = a i j ,   i ,   j ∈ 1 , 2 , 3 , be the 3 × 3 matrix such that a i j = 1 if j + 1 is divisible by i , otherwise a i j = 0 . Then which of the following statements is(are) true?
JEE Advanced
Mathematics
Matrices
2023
JEE Advanced 2023 (Paper 2)
JEE Advanced Mathematics Question (2023) — Solution
Options
- A. M is invertible
- B. There exists a nonzero column matrix a 1 a 2 a 3 such that M a 1 a 2 a 3 = - a 1 - a 2 - a 3
- C. The set X ∈ ℝ 3 : M X = 0 ≠ 0 , where 0 = 0 0 0
- D. The matrix ( M - 2   I ) is invertible, where I is the 3 × 3 identity matrix
Answer
C. The set X ∈ ℝ 3 : M X = 0 ≠ 0 , where 0 = 0 0 0
Step-by-step solution
Given, M = a i j ,   i ,   j ∈ 1 , 2 , 3 a i j = 1  if  j + 1  is divisible by  i otherwise  a i j = 0 So, a 11 = 1 ,   a 12 = 1 ,   a 13 = 1 ,   a 21 = 1 ,   a 22 = 0 . . . . . . , a 33 = 0 So, M = 1      1      1 1      0      1 0      1      0 ⇒ | M | = 1 ( − 1 ) − 1 ( − 1 ) = − 1 + 1 = 0 Hence, M is not invertible Now solving option B we get, 1 1 1 1 0 1 0 1 0 a 1 a 2 a 3 = - a 1 - a 2 - a 3 ⇒ a 1 + a 2 + a 3 a 1 + a 3 a 2 = - a 1 - a 2 - a 3 Now on comparing we get, a 1 + a 2 + a 3 = - a 1 ,   a 1 + a 3 = - a 2   &   a 2 = - a 3 Now on solving we get, a 1 = 0   &   a 2 + a 3 = 0 So, there will be infinite possibility for a 2   &   a 3 ⇒ There exist a column matrix (infinite possibilities) Now solving option C we get, 1 1 1 1 0 1 0 1 0 x y z = 0 0 0 ⇒ x + y + z x + z y = 0 0 0 Now on comparing we get, ⇒ x + y + z = 0 x + z = 0 y = 0 Yes it is possible Now solving option D we get, | M − 2 I | = − 1 1 1 1 − 2 1 0 1 − 2 ⇒ | M − 2 I | = − 1 3 − 1 − 2 − 1 = − 3 + 3 = 0 Hence, it is not invertible.
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