Question

Say true or false:
If x = by + cz, y = az + cx, z = bx + ay, where x, y, z are not all zero, then ${{\text{a}}^{2}}\text{ + }{{\text{b}}^{2}}\text{ + }{{\text{c}}^{2}}\text{ + 2abc + 1 = 0}$
A. True
B. False

Answer 1

Hint: Substitute any of the variables, say z, in two equations in the form of the third equation and express the other two variables, x and y in terms of ratio for the two equations separately. Now equate the ratios x:y for the two equations and find the relation between the coefficients a, b, c. Thus, find out whether the given statement in the question is true or false.

Complete step-by-step answer:
The three given equations are as follows:
x = by + cz ..... (i)
y = az + cx ...... (ii)
z = ax + by ......(iii)
Thus replacing z = bx + ay from the equation (iii) in the equation (i) and (ii) respectively, we get,
\[\begin{align}
  & \text{x = by + c}\cdot \left( \text{bx + ay} \right)\text{ }....\text{(iv)} \\
 & \text{y = a}\cdot \left( \text{bx + ay} \right)\text{ + cx }....\left( \text{v} \right) \\
\end{align}\]
Simplifying equation (iv), we obtain the ratio of x and y as follows:
\[\begin{align}
  & \text{x = by + bcx + acy} \\
 & \Rightarrow \text{ x}\left( 1\text{ }-\text{ bc} \right)\text{ = y}\left( \text{b + ac} \right) \\
 & \therefore \text{ x:y = }\left( \text{b + ac} \right):\left( 1\text{ }-\text{ bc} \right)\text{ }....\left( \text{A} \right) \\
\end{align}\]
Similarly, by simplifying equation (v), we obtain the ratio of x and y as follows:
\[\begin{align}
  & \text{y = abx + }{{\text{a}}^{2}}\text{y}\,\text{+ cx } \\
 & \Rightarrow \text{ x}\left( \text{ab + c} \right)\text{ = y}\left( \text{1 }-\text{ }{{\text{a}}^{2}} \right) \\
 & \therefore \text{ x:y = }\left( \text{1 }-\text{ }{{\text{a}}^{2}} \right):\left( \text{ab + c} \right)\text{ }....\left( \text{A} \right) \\
\end{align}\]
Thus, from both the sets of ratios of x and y in (A) and (B), we can equate them and thus find the relationship between a, b and c as follows:
$\begin{align}
  & \left( \text{b + ac} \right):\left( 1\text{ }-\text{ bc} \right)\text{ = }\left( 1\text{ }-\text{ }{{\text{a}}^{2}} \right):\left( \text{ab + c} \right) \\
 & \Rightarrow \text{ }\left( \text{b + ac} \right)\cdot \left( \text{ab + c} \right)\text{ = }\left( 1\text{ }-\text{ }{{\text{a}}^{2}} \right)\cdot \left( 1\text{ }-\text{ bc} \right) \\
 & \Rightarrow \text{ a}{{\text{b}}^{2}}\text{ + bc + }{{\text{a}}^{2}}\text{bc + a}{{\text{c}}^{2}}\text{ = 1 }-\text{ }{{\text{a}}^{2}}-\text{ bc + }{{\text{a}}^{2}}\text{bc} \\
 & \therefore \text{ }{{\text{a}}^{2}}\text{ + a}{{\text{b}}^{2}}\text{ }+\text{ a}{{\text{c}}^{2}}\text{ + 2bc }-\text{ 1 = 0} \\
\end{align}$
Thus, we can very easily observe that the relation between the coefficients a, b and c does not turn out to be the statement given in the question. Hence, we can say that the statement in the question is false.
Thus, the correct answer is option B.

Note: The variables x and y can be expressed in the term of ratios only because it is given in the question that x, y and z are not all zero. Otherwise, they could not be solved in the above method. This problem can alternatively be solved by the matrix method. The three equations can be expressed in the form of three homogeneous equations. We prepare a 3x3 matrix with the coefficients of x, y and z and (0,0,0) is the trivial solution of the system of equations. Since, it is given that all are not equal to zero, we can understand there are infinitely many solutions and thus, the determinant of the coefficient matrix will be equal to zero. Equating the determinant to zero as well, we obtain the same relation ${{\text{a}}^{2}}\text{ + a}{{\text{b}}^{2}}\text{ }+\text{ a}{{\text{c}}^{2}}\text{ + 2bc }-\text{ 1 = 0}$.