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JEE Main 2018

JEE Main 2018 Mathematics question on Matrices and Determinants

If ∣x−42x2x2xx−42x2x2xx−4∣=(A+Bx)(x−A)2\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix} = (A + Bx)(x - A)^2, then the ordered pair (A,B)(A, B) is equal to
  1. A.
    (−4,−5)(-4, -5)
  2. B.
    (−4,3)(-4, 3)
  3. C.
    (−4,5)(-4, 5)
  4. D.
    (4,5)(4, 5)
Show answer and solution

Correct option: C

Solution

Apply R1 → R1 + R2 + R3 to get the first row as (5x-4, 5x-4, 5x-4) = (5x-4)(1,1,1).

Factor out (5x-4), then subtract C1 from C2 and C1 from C3 to get a triangular-like matrix. After column operations C2←C2-C1, C3←C3-C1, the matrix becomes diagonal with (5x-4) factor and two rows giving (x-4-2x) = -(x+4) on the diagonal.

So the determinant = (5x-4)(-(x+4))² = (5x-4)(x+4)².

Matching with (A+Bx)(x-A)²: we need x-A = x+4, so A = -4, and A+Bx = 5x-4 = -4+5x, so B = 5.

The ordered pair is (-4, 5).

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