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Matrices and Determinants question

2019 · 10 Jan · Shift 2 · Q31
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Matrices and Determinants question

2019 · 10 Jan · Shift 2 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A = [2b1bb2+1b1b2]\left[ {\begin{matrix} 2 & b & 1 \\ b & {{b^2} + 1} & b \\ 1 & b & 2 \\ \end{matrix} } \right]​2b1​bb2+1b​1b2​​ where b > 0. Then the minimum value of det⁡(A)b{{\det \left( A \right)} \over b}bdet(A)​ is -
  1. A
    3\sqrt 33​
  2. B
    −23-2\sqrt 3−23​
  3. C
    −3- \sqrt 3−3​
  4. D
    232\sqrt 323​
View written solutionFree

Correct answer: D

  1. We need to find the minimum value of
det⁡(A)b,b>0,\frac{\det(A)}{b}, \qquad b>0,bdet(A)​,b>0,

where

A=[2b1bb2+1b1b2].A=\begin{bmatrix} 2 & b & 1\\ b & b^2+1 & b\\ 1 & b & 2 \end{bmatrix}.A=​2b1​bb2+1b​1b2​​.
  1. Compute det⁡(A)\det(A)det(A). Using expansion along the first row:
det⁡(A)=2∣b2+1bb2∣−b∣bb12∣+1∣bb2+11b∣.\det(A)=2\begin{vmatrix} b^2+1 & b \\ b & 2 \end{vmatrix} -b\begin{vmatrix} b & b \\ 1 & 2 \end{vmatrix} +1\begin{vmatrix} b & b^2+1 \\ 1 & b \end{vmatrix}.det(A)=2​b2+1b​b2​​−b​b1​b2​​+1​b1​b2+1b​​.

Now evaluate each minor:

∣b2+1bb2∣=2(b2+1)−b2=b2+2,\begin{vmatrix} b^2+1 & b \\ b & 2 \end{vmatrix}=2(b^2+1)-b^2=b^2+2,​b2+1b​b2​​=2(b2+1)−b2=b2+2, ∣bb12∣=2b−b=b,\begin{vmatrix} b & b \\ 1 & 2 \end{vmatrix}=2b-b=b,​b1​b2​​=2b−b=b, ∣bb2+11b∣=b2−(b2+1)=−1.\begin{vmatrix} b & b^2+1 \\ 1 & b \end{vmatrix}=b^2-(b^2+1)=-1.​b1​b2+1b​​=b2−(b2+1)=−1.

Therefore,

det⁡(A)=2(b2+2)−b(b)+(−1)=2b2+4−b2−1=b2+3.\det(A)=2(b^2+2)-b(b)+(-1)=2b^2+4-b^2-1=b^2+3.det(A)=2(b2+2)−b(b)+(−1)=2b2+4−b2−1=b2+3.
  1. Hence,
det⁡(A)b=b2+3b=b+3b,b>0.\frac{\det(A)}{b}=\frac{b^2+3}{b}=b+\frac{3}{b}, \qquad b>0.bdet(A)​=bb2+3​=b+b3​,b>0.
  1. Now minimize
f(b)=b+3b,b>0.f(b)=b+\frac{3}{b}, \qquad b>0.f(b)=b+b3​,b>0.

Using AM-GM:

b+3b≥2b⋅3b=23.b+\frac{3}{b} \ge 2\sqrt{b\cdot \frac{3}{b}}=2\sqrt{3}.b+b3​≥2b⋅b3​​=23​.

Equality holds when

b=3b  ⟹  b2=3  ⟹  b=3b=\frac{3}{b} \implies b^2=3 \implies b=\sqrt{3}b=b3​⟹b2=3⟹b=3​

(since b>0b>0b>0).

  1. Therefore the minimum value is
23.2\sqrt{3}.23​.
  1. Comparing with the options:
  • A: 3\sqrt33​
  • B: −23-2\sqrt3−23​
  • C: −3-\sqrt3−3​
  • D: 232\sqrt323​

So the correct option is D.

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