Matrices And Determinants Ques 1
- Let $A=\left[\begin{array}{ccc}2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2\end{array}\right]$, where $b>0$. Then, the minimum value of $\frac{\operatorname{det}(A)}{b}$ is
(2019 Main, 10 Jan II)
(a) $-\sqrt{3}$
(b) $-2 \sqrt{3}$
(c) $2 \sqrt{3}$
(d) $\sqrt{3}$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Given matrix, $A=\left[\begin{array}{ccc}2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2\end{array}\right], b>0$
$ \begin{aligned} \text { So, } \operatorname{det}(A)=|A| & =\left|\begin{array}{ccc} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{array}\right| \\ & =2\left[2\left(b^2+1\right)-b^2\right]-b(2 b-b) \\ & =2\left[2 b^2+2-b^2\right]-b^2-1+1\left(b^2-b^2-1\right) \\ & =2 b^2+4-b^2-1=b^2+3 \\ \Rightarrow \quad \frac{\operatorname{det}(A)}{b} & =\frac{b^2+3}{b}=b+\frac{3}{b} \end{aligned} $
Now, by $\mathrm{AM} \geq \mathrm{GM}$, we get
$ \begin{aligned} & \frac{b+\frac{3}{b}}{2} \geq\left(b \times \frac{3}{b}\right)^{1 / 2} \\ & \{\because b>0\} \\ & \Rightarrow \quad b+\frac{3}{b} \geq 2 \sqrt{3} \\ \end{aligned} $
So, minimum value of $\frac{\operatorname{det}(A)}{b}=2 \sqrt{3}$
BETA
