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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Matrices and Determinants

Let A=[2b1bb2+1b1b2]A = \begin{bmatrix} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{bmatrix}, where b>0b > 0. Then, the minimum value of det⁡(A)b\dfrac{\det(A)}{b} is
  1. A.
    −3-\sqrt{3}
  2. B.
    −23-2\sqrt{3}
  3. C.
    232\sqrt{3}
  4. D.
    3\sqrt{3}
Show answer and solution

Correct option: C

Solution

Compute det(A) by expanding. Using cofactor expansion or direct calculation:

det(A) = 2[(b²+1)·2 - b·b] - b[b·2 - b·1] + 1[b·b - (b²+1)·1]

= 2[2b²+2 - b²] - b[2b - b] + 1[b² - b²-1]

= 2[b²+2] - b[b] + (-1)

= 2b²+4 - b² - 1

= b² + 3

So det(A)/b = (b² + 3)/b = b + 3/b.

Since b > 0, by AM-GM: b + 3/b ≥ 2√3, with equality when b = √3.

The minimum value is 2√3.

[JEE Main 2019 · 10 Jan · Shift II]

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