Let A = [2&b&1 b&b2+1&b 1&b&2] where b 0. Then the minimum value of (det (A)/b) is :

Question

Let A = [2&b&1 b&b2+1&b 1&b&2] where b > 0. Then the minimum value of (det (A)/b) is : (A) √3 (B)  - √3 (C)  - 2 √3 (D) 2 √3

Tardigrade Admin · Accepted Answer

A = [2&b&1 b&b2+1&b 1&b&2] (b>0) |A| = 2(2b2 +2 -b2 )-b(2b-b)+1(b2-b2-1) |A| =2(b2+2)-b2-1 |A|=b2+3 (|A|/b) =b+(3/b) ⇒ (b+ (3/b)/2) ge√3 b+ (3/b) ge2√3

Q. Let $A = ⎣ ⎡ 2 b 1 b b^{2} + 1 b 1 b 2 ⎦ ⎤$ where $b > 0$ . Then the minimum value of $\frac{d e t ( A )}{b}$ is :

$3$

$- 3$

$- 23$

$23$

Q. Let A=⎣⎡​2b1​bb2+1b​1b2​⎦⎤​ where b>0. Then the minimum value of bdet(A)​ is :

3​

−3​

−23​

23​

A=⎣⎡​2b1​bb2+1b​1b2​⎦⎤​(b>0) ∣A∣=2(2b2+2−b2)−b(2b−b)+1(b2−b2−1) ∣A∣=2(b2+2)−b2−1 ∣A∣=b2+3 b∣A∣​=b+b3​⇒2b+b3​​≥3​ b+b3​≥23​

Q. Let $A = ⎣ ⎡ 2 b 1 b b^{2} + 1 b 1 b 2 ⎦ ⎤$ where $b > 0$ . Then the minimum value of $\frac{d e t ( A )}{b}$ is :

$3$

$- 3$

$- 23$

$23$