Question

Let $A=\left[ \begin{matrix}
   2 & b & 1 \\
   b & {{b}^{2}}+1 & b \\
   1 & b & 2 \\
\end{matrix} \right]$ where $b>0$. Then the minimum value of $\dfrac{\det \left( A \right)}{b}$ is :
A. $\sqrt{3}$
B. $-\sqrt{3}$
C. $-2\sqrt{3}$
D. $2\sqrt{3}$

Answer 1

Hint: Here in the given matrix of $3\times 3$ ratio we will first find out the value of its determinant and as we can see, it might be obtained in variable b. Then to find the minimum value of $\dfrac{\det \left( A \right)}{b}$, we use the concept of maxima - minima and carry out the derivative test to find the minimum value.

Complete step-by-step answer:
Here in the question we have been given a $3\times 3$ matrix, which means that the matrix has 3 rows and 3 columns and it is categorised under square matrix as per the type of matrices. And we have been asked to find the value of $\dfrac{\det \left( A \right)}{b}$. So, we will first find the determinant of matrix A, which would be,
$\det \left( A \right)=\left| \begin{matrix}
   2 & b & 1 \\
   b & {{b}^{2}}+1 & b \\
   1 & b & 2 \\
\end{matrix} \right|$
We will now expand the determinal along row 1. So, we will get,
$\begin{align}
  & 2\left[ 2\left( {{b}^{2}}+1 \right)-{{b}^{2}} \right]-b\left[ 2b-b \right]+1\left[ {{b}^{2}}-\left( {{b}^{2}}+1 \right) \right] \\
 & \Rightarrow 2\left[ 2{{b}^{2}}+2-{{b}^{2}} \right]-b\left( b \right)+1\left( {{b}^{2}}-{{b}^{2}}-1 \right) \\
 & \Rightarrow 2\left[ {{b}^{2}}+2 \right]-{{b}^{2}}-1 \\
 & \Rightarrow 2{{b}^{2}}+4-{{b}^{2}}-1 \\
 & \Rightarrow {{b}^{2}}+3 \\
\end{align}$
Now, we have got the simplified form of the determinant, ${{b}^{2}}+3$. So, let us take $f\left( b \right)=\dfrac{{{b}^{2}}+3}{b}$. This is the function of b whose minimum value has to be found. So, we will differentiate $f\left( b \right)$ and then find the value of b, where $f'\left( b \right)=0$ and then find the value at which $f\left( b \right)$ is minimum. So, on differentiating, we will get,
$f'\left( b \right)=1-\dfrac{3}{{{b}^{2}}}$
On equating $f'\left( b \right)=0$, we get,
$\begin{align}
  & 1-\dfrac{3}{{{b}^{2}}}=0 \\
 & \Rightarrow \dfrac{{{b}^{2}}-3}{{{b}^{2}}}=0 \\
 & \Rightarrow {{b}^{2}}-3=0 \\
 & \Rightarrow {{b}^{2}}=3 \\
 & \Rightarrow b=\pm \sqrt{3} \\
\end{align}$
Now, we will substitute the value of $b=\sqrt{3}$ in the $f\left( b \right)$, so we get the value as,
$\begin{align}
  & f\left( b \right)=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+3}{\sqrt{3}} \\
 & \Rightarrow f\left( b \right)=\dfrac{3+3}{\sqrt{3}} \\
 & \Rightarrow f\left( b \right)=\dfrac{6}{\sqrt{3}}=2\sqrt{3} \\
\end{align}$
Now, we will take the value of $b=-\sqrt{3}$, so we get the value of $f\left( b \right)$ as,
$\begin{align}
  & f\left( b \right)=\dfrac{{{\left( -\sqrt{3} \right)}^{2}}+3}{-\sqrt{3}} \\
 & \Rightarrow f\left( b \right)=-\dfrac{3+3}{\sqrt{3}} \\
 & \Rightarrow f\left( b \right)=-\dfrac{6}{\sqrt{3}}=-2\sqrt{3} \\
\end{align}$
Thus, we get the minimum value as $-2\sqrt{3}$.

So, the correct answer is “Option c”.

Note: We can use an alternate method for solving the determinant. To find the minimum value of the function of b, that is $\dfrac{\det \left( A \right)}{b}$, the derivative test can be used or we can make a quadratic equation in b and put the value of $D\ge 0$ and find the range of $f\left( b \right)$ and check for the minimum value. The students should be careful with their calculations as this involves a lot of multiple operations and there are chances of errors with numbers or signs.