Let A = 2 b 1 b b 2 + 1 b 1 b 2 , where b > 0 . Then the minimum value of det A b is:

A = 2 b 1 b b 2 + 1 b 1 b 2 = 2 2 b 2 + 2 - b 2 - b 2 b - b + 1 b 2 - b 2 - 1 = 2 b 2 + 2 - b 2 - 1 = 2 b 2 + 4 - b 2 - 1 = b 2 + 3 A b = b + 3 b = b - 3 b 2 + 2 3 Hence, the minimum value of A b is 2 3 . ∵ b - 3 b 2 ≥ 0

Mathematics - Determinant Question with Solution | TestHub

MathematicsDeterminantProperties of det.Medium2 minPYQ_2019

MathematicsMediumsingle choice

Let $A = [\begin{matrix} 2 & b & 1 \\ b & b^{2} + 1 & b \\ 1 & b & 2 \end{matrix}],$ where $b > 0$ . Then the minimum value of $\frac{\det (A)}{b}$ is:

Options:

Answer:

Solution:

$|A| = |\begin{matrix} 2 & b & 1 \\ b & b^{2} + 1 & b \\ 1 & b & 2 \end{matrix}|$

$= 2 (2 b^{2} + 2 - b^{2}) - b (2 b - b) + 1 (b^{2} - b^{2} - 1)$

$= 2 (b^{2} + 2) - b^{2} - 1$

$= 2 b^{2} + 4 - b^{2} - 1$

$= b^{2} + 3$

$\frac{|A|}{b} = b + \frac{3}{b} = {(\sqrt{b} - \sqrt{\frac{3}{b}})}^{2} + 2 \sqrt{3}$

Hence, the minimum value of $\frac{|A|}{b}$ is $2 \sqrt{3}$ . $(∵ {(\sqrt{b} - \sqrt{\frac{3}{b}})}^{2} \geq 0)$

Stream:JEESubject:MathematicsTopic:DeterminantSubtopic:Properties of det.

⏱ 2mℹ️ Source: PYQ_2019

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