Question
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If a set of linear equations given as: $x=cy+bz,y=cx+az,z=bx+ay$ has a non-trivial solution, then value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc$ equals to
a) 0
b) -1
c) 1
d) None of these

Answer Verified Verified
Hint: We have following equations:
$\begin{align}
  & x=cy+bz......(1) \\
 & y=cx+az......(2) \\
 & z=bx+ay......(3) \\
\end{align}$
Since it is said that the set of the given nonhomogeneous equations has a non-trivial solution, it is only possible if the determinant is zero.
So, after rearranging the given set of linear equations in the form of $ax+by+cz=0$, we can fill up the elements of the determinant as follows:
$\Delta =\left| \begin{matrix}
   -1 & c & b \\
   c & -1 & a \\
   b & a & -1 \\
\end{matrix} \right|$
Since, the given set of linear equations has a non-trivial solution, we can write:$\left| \begin{matrix}
   -1 & c & b \\
   c & -1 & a \\
   b & a & -1 \\
\end{matrix} \right|=0$
Therefore, find the determinant of the given equations and put it equal to zero. Thereby, get the equation in terms of a, b and c as required and solve to get the equation in the required form, i.e. ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc$ .

Complete step by step answer:
Since, we have the following equations:
$\begin{align}
  & x=cy+bz......(1) \\
 & y=cx+az......(2) \\
 & z=bx+ay......(3) \\
\end{align}$
We can also write the given equations as:
$\begin{align}
  & -x+cy+bz=0......(4) \\
 & cx-y+az=0......(5) \\
 & bx+ay-z=0......(6) \\
\end{align}$

So, we can say that the given equations will have a non-trivial solution if and only if their determinant is zero, i.e.
$\left| \begin{matrix}
   -1 & c & b \\
   c & -1 & a \\
   b & a & -1 \\
\end{matrix} \right|=0$
Now, by solving the above determinant, we get:
$\begin{align}
  & -1\left( 1-{{a}^{2}} \right)-c\left( -c-ab \right)+b\left( ac+b \right)=0 \\
 & -1+{{a}^{2}}+{{c}^{2}}+abc+abc+{{b}^{2}}=0 \\
 & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc-1=0 \\
 & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc=1 \\
\end{align}$

Hence, we get the desired equation and the result as:
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc=1$

So, the correct answer is “Option C”.

Note: While rearranging the equation, take care of the sign of the variables in the equation. A single change in sign might change the whole answer. Also, while solving the determinant, be careful for the same reason. You can also use shortcuts to solve the determinants.
Always remember that a set of equations have a non-trivial solution if and only if the determinant is equal to zero. If the determinant is non-zero, the equations might have an infinite number of solutions. Hence, we might not be able to get the required equation as well as the result.