Question

The system of equations ax + y + z = 0, x + by + z = 0, x + y + cz = 0 has a non-trivial solution then $\dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = $

Answer 1

Hint: In this question this system of equations has a non-trivial solution then it means it has non-zero solutions. We will solve the given equations and find $\dfrac{1}{{1 - a}}, \dfrac{1}{{1 - b}}, \dfrac{1}{{1 - c}}$ in terms of x, y, z. Then finally add these three equations to get the result.

Complete step-by-step answer:
We have;
ax + y + z =0 $ \Rightarrow $ y + z = − ax
Adding x on both sides;
x + y + z = -ax + x $ \Rightarrow $ x + y + z = x (1 – a)
$ \Rightarrow \dfrac{1}{{1 - a}} = \dfrac{x}{{x + y + z}}$ …(i)
Again, x + by + z = 0 $ \Rightarrow $x + z = − by
Adding y on both sides;
x + y + z = -by + y $ \Rightarrow $x + y + z = y (1 – b)
$ \Rightarrow \dfrac{1}{{1 - b}} = \dfrac{y}{{x + y + z}}$ …(ii)
Again, we have;
x + y + cz = 0 $ \Rightarrow $x + y = − cz
Adding z on both sides
x + y + z = − cz + z $ \Rightarrow $x + y + z = z (1 – c)
$ \Rightarrow \dfrac{1}{{1 - c}} = \dfrac{z}{{x + y + z}}$ …(iii)
Adding equations (i), (ii), and (iii), we get;
$\dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = \dfrac{x}{{x + y + z}} + \dfrac{y}{{x + y + z}} + \dfrac{z}{{x + y + z}}$
$ \Rightarrow \dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = \dfrac{{x + y + z}}{{x + y + z}}$
$ \Rightarrow \dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = 1$.

Hence, $\dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = $1.

Note: The system of equations in which the determinant of the coefficient is zero is called a non-trivial solution. And the system of equations in which the determinant of the coefficient matrix is not zero but the solution x = y = z = 0 is called a trivial solution.
Alternatively, we can solve this question, using matrix rule. We have three equations and three variables, first write the system of equations in matrix format. And find the values following the steps.