Answer
Verified
396.3k+ views
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.
$\begin{align}
& x=cy+bz......(1) \\
& y=cx+az......(2) \\
& z=bx+ay......(3) \\
\end{align}$
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE