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JEE Main 2016

JEE Main 2016 Mathematics question on Matrices and Determinants

If A=[5a−b32]A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} and AA adj A=AATA = AA^T, then 5a+b5a + b is equal to
  1. A.
    −1-1
  2. B.
    55
  3. C.
    44
  4. D.
    1313
Show answer and solution

Correct option: B

Solution

For any matrix A, A·adj(A) = det(A)·I. Also, AAT = det(A)·I only when A is such that the given condition holds, meaning det(A)·I = AAT.

Computing det(A) = 10a + 3b.

For AAT = det(A)·I, comparing entries: the (1,2) entry of AAT must be 0, and diagonal entries must equal det(A).

AAT (1,2) entry = (5a)(3) + (-b)(2) = 15a - 2b = 0, so b = 15a/2.

AAT (1,1) entry = 25a² + b² = det(A) = 10a + 3b.

Substituting b = 15a/2: 25a² + 225a²/4 = 10a + 45a/2
→ (100a² + 225a²)/4 = (20a + 45a)/2
→ 325a²/4 = 65a/2
→ 325a = 130
→ a = 2/5, and b = 15(2/5)/2 = 3.

Therefore 5a + b = 5(2/5) + 3 = 2 + 3 = 5.

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