Question

Let A = where b > 0. Then the minimum value of det(A)/b
is :
(A) 2√3 (B) –2√3
(C) –√3 (D) √3

Answer 1

Step-by-step explanation:

Given Given Let A = where b > 0. Then the minimum value of det(A)/b

is  

• Given A = [2             b             1
•                   b         b^2 + 1       b
•                   1            b              2]
• Now det (A) = 2(2b^2 + 2 – b^2) – b(2b – b) + 1(b^2 – b^2 – 1)
•                    = 2b^2 + 4 – b^2 – 1
•                   = b^2 + 3
• Now minimum value of det (A) / b = b^2 + 3/b
•                                                       = b + 3/b
• Differentiating we get  f’ (b) = 1 – 3/b^2 = 0
•                                            Or b^2 = 3
•                                             Or b = √3
•          Differentiating again we get f’(b) = 0 – (-2) x 3 / b^3
•                                                Now 6 / b^3 > 0
• So minimum value det (A) / b = b^2 + 3 / b
•          = (√3)^2 + 3 / √3
•           = 6 / √3
•          = 2 √3

Reference link will be

https://brainly.in/question/11474862

Answer 2

Given:

$\quad\quad A=\left[\begin{array}{ccc} 2 & b & 1\\ b & b^{2}+1 & b\\ 1 & b & 2 \end{array}\right]$

Solution:

Then, det(A)

$\quad=\begin{array}{|ccc|} 2 & b & 1\\ b & b^{2}+1 & b\\ 1 & b & 2\end{array}$

Expanding along the first row, we get:

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

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

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

$\quad=b^{2}+3$

To attain the minimum value of $\frac{det(A)}{b}$:

Now, $\frac{det(A)}{b}$

$\quad=\frac{b^{2}+3}{b}$

$\quad=b+\frac{3}{b}$

$\to f(b)=b+\frac{3}{b}$

$\to f'(b)=1-\frac{3}{b^{2}}$

To attain the minimum value, we take $f'(b)=0$:

$\quad 1-\frac{3}{b^{2}}=0$

$\to b=\sqrt{3}$

∴ the minimum value of $\frac{det(A)}{b}$ is

$\quad=\frac{(\sqrt{3})^{2}+3}{\sqrt{3}}$

$\quad\quad=\frac{6}{\sqrt{3}}$

$\quad\quad\quad=2\sqrt{3}$

Option (A) is correct.

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