Question

If $a,b,c$ are sides of scalene triangle, then the value of $\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|$ is
A. Non-negative
B. negative
C. positive
D. Non-positive

Answer 1

We are given three sides of scalene triangle which are $a,b,c$. Also, we have given a matrix $\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|$ . Find the value of $\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|$ and apply the rules of the scalene triangle. Try it, you will get the answer.

Complete step-by-step answer:
Now we are given sides of the scalene triangle which are $a,b,c$.
We know the conditions for the scalene triangle that if $a,b,c$ are sides of the scalene triangle then the sides are greater than zero.
So we can say in mathematical form,
$a,b,c>0$
Also, since the triangle is scalene we know that all its sides are not equal.
In mathematical form we can write it as,
$a\ne b\ne c$
Now let us take the matrix which is given in the above problem and let us find its determinant.
The matrix we are given is $\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|$.
So solving above matrix we get,
$\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|=a(bc-{{a}^{2}})-b({{b}^{2}}-ac)+c(ab-{{c}^{2}})$
Now simplifying in simple form we get,
$\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|=abc-{{a}^{3}}-{{b}^{3}}+abc+abc-{{c}^{3}}$
$\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|=3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}$
So we have got the value of matrix $\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|$ as $3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}$.
Now let us check whether the value of the matrix is negative or positive or anything from the options.
Now let us consider the numbers ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ and $a,b,c>0$ i.e. $a,b,c$ are positive, so let us use A.M$\ge$G.M.
Now A.M i.e. arithmetic mean will be $\dfrac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{3}$.
Also, G.M i.e. geometric mean will be ${{({{a}^{3}}{{b}^{3}}{{c}^{3}})}^{\dfrac{1}{3}}}$.
So using A.M$\ge$G.M and substituting the values we get.
$\dfrac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{3}\ge {{({{a}^{3}}{{b}^{3}}{{c}^{3}})}^{\dfrac{1}{3}}}$
Now simplifying above we get and multiplying above equation by $3$, we get,
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}\ge 3(abc)$
Now again simplifying in simple manner we get,
$3(abc)-({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\le 0$
So since $3(abc)-({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\le 0$ we can say that it is negative.
Therefore, if $a,b,c$ are sides of scalene triangle, then the value of $\left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & b \\\ \end{matrix} \right|$ is negative.
So the correct answer is option (B).

Note: Here you should know A.M i.e. arithmetic mean and G.M i.e. geometric mean. There should be proper considerations made as in problem we can see that ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ is considered. You must know the property that if $a,b,c>0$ i.e. $a,b,c$ are positive then A.M$\ge$G.M.