Question

If the value of the determinant $\left| \begin{matrix} a & 1 & 1 \\\ 1 & b & 1 \\\ 1 & 1 & c \\\ \end{matrix} \right|$ is positive, then $\left( a,b,c>0 \right)$
A) $abc>1$
B) $abc>-8$
C) $abc < -8$

Answer 1

Here in this question we have been given that the value of the determinant $\left| \begin{matrix} a & 1 & 1 \\\ 1 & b & 1 \\\ 1 & 1 & c \\\ \end{matrix} \right|$ is positive. We will find the determinant and then we will simplify it using the simple logic given as that for any three numbers $a,b,c$ the expression $\dfrac{a+b+c}{2}\ge {{\left( abc \right)}^{\dfrac{1}{2}}}\ge \dfrac{3abc}{ab+bc+ca}$ is valid.

Complete step by step solution:
Now considering from the question we have been asked to find the expression for $abc$ when the value of the determinant $\left| \begin{matrix} a & 1 & 1 \\\ 1 & b & 1 \\\ 1 & 1 & c \\\ \end{matrix} \right|$ is positive.
From the basic concepts of progressions we know that for $n$ numbers $\text{Arithmetic mean }\ge \text{ Geometric mean }\ge \text{ Harmonic mean}$ . The arithmetic mean of $n$ numbers is given as $\dfrac{{{a}_{1}}+{{a}_{2}}+.........+{{a}_{n}}}{n}$ , geometric mean is given as ${{\left( {{a}_{1}}{{a}_{2}}......{{a}_{n}} \right)}^{\dfrac{1}{n}}}$ and harmonic mean is given as $\dfrac{n}{\dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+......+\dfrac{1}{{{a}_{n}}}}$ .
Now if we consider $a,b,c$ then we will have $\dfrac{a+b+c}{3}\ge {{\left( abc \right)}^{\dfrac{1}{3}}}$ .
Now we will compute the determinant of the matrix $\left| \begin{matrix} a & 1 & 1 \\\ 1 & b & 1 \\\ 1 & 1 & c \\\ \end{matrix} \right|$ is given as $\begin{aligned} & \Rightarrow a\left| \begin{matrix} b & 1 \\\ 1 & c \\\ \end{matrix} \right|-1\left| \begin{matrix} 1 & 1 \\\ c & 1 \\\ \end{matrix} \right|+1\left| \begin{matrix} 1 & b \\\ 1 & 1 \\\ \end{matrix} \right|=a\left( bc-1 \right)-\left( c-1 \right)+\left( 1-b \right) \\\ & \Rightarrow abc-a-b-c+2 \\\ \end{aligned}$
Now from the question we know that the determinant is positive so $abc-a-b-c+2>0$ .
Now we can further simplify it using the relation $\dfrac{a+b+c}{3}\ge {{\left( abc \right)}^{\dfrac{1}{3}}}$ and write it as $\begin{aligned} & \Rightarrow abc+2-\left( a+b+c \right)>0 \\\ & \Rightarrow abc+2>\left( a+b+c \right) \\\ & \Rightarrow abc+2>3{{\left( abc \right)}^{\dfrac{1}{3}}} \\\ \end{aligned}$
Let us assume that ${{\left( abc \right)}^{\dfrac{1}{3}}}=x$ then we can write it as
$\begin{aligned} & {{x}^{3}}+2>3x \\\ & \Rightarrow {{x}^{3}}-3x+2>0 \\\ & \Rightarrow {{\left( x-1 \right)}^{2}}\left( x+2 \right)>0 \\\ \end{aligned}$
As ${{\left( x-1 \right)}^{2}}$ is always positive so if $\left( x+2 \right)>0$ then the expression is valid so $\begin{aligned} & x>-2\Rightarrow {{\left( abc \right)}^{\dfrac{1}{3}}}>-2 \\\ & \Rightarrow abc>{{\left( -2 \right)}^{3}} \\\ & \Rightarrow abc>-8 \\\ \end{aligned}$ .
Hence we can conclude that the value of the determinant $\left| \begin{matrix} a & 1 & 1 \\\ 1 & b & 1 \\\ 1 & 1 & c \\\ \end{matrix} \right|$ is positive, then $abc>-8$ .

So, the correct answer is “Option B”.

Note: While answering questions of this type we should be sure with our algebraic concepts. This question can be answered easily and in a short span of time and very few mistakes are possible in it. Similarly we have another relation between arithmetic, geometric and harmonic means which is mathematically given as $\text{Geometric Mean = }\sqrt{\text{Arithmetic Mean}\times \text{Harmonic Mean}}$.