Questions & Answers

Question

Answers

a) 0

b) -1

c) 1

d) None of these

Answer
Verified

$\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. ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc$ .

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:

${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc=1$

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.