Question

The system of linear equations:
$x + \lambda y - z = 0$
$\lambda x - y - z = 0$
$x + y - \lambda z = 0$
Has a non-trivial solution for:
A. infinitely many values of $\lambda$
B. exactly one value of $\lambda$
C. exactly two values of $\lambda$
D. exactly three values of $\lambda$

Answer 1

If we are given the system of equations like here we are given:
$x + \lambda y - z = 0$
$\lambda x - y - z = 0$
$x + y - \lambda z = 0$
If all variable values are not equal to $0$, then it is a non-trivial solution.
\left| {\begin{array}{*{20}{c}} 1&\lambda &{ - 1} \\\ \lambda &{ - 1}&{ - 1} \\\ 1&1&{ - \lambda } \end{array}} \right| = 0
Get the determinant of coefficients and find the value of $\lambda$.

Complete step-by-step answer:
So here we are given the system of equations like here we are given:
$x + \lambda y - z = 0$
$\lambda x - y - z = 0$
$x + y - \lambda z = 0$
And it is a non-trivial solution.
It means that at least one of the variables must not be equal to $0$.
If all are equal to zero, then it is a trivial solution.
So here the determinant of the coefficients as shown must be equal to $0$
\left| {\begin{array}{*{20}{c}} 1&\lambda &{ - 1} \\\ \lambda &{ - 1}&{ - 1} \\\ 1&1&{ - \lambda } \end{array}} \right| = 0
Now solving it, we get
$1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0$
$(\lambda + 1) - \lambda (1 - {\lambda ^2}) - 1(\lambda + 1) = 0$
We know identity $(a-b) (a+b) = a^2-b^2$
Applying the identity to 2nd term
$(\lambda + 1) + \lambda ({\lambda ^2}-1) - 1(\lambda + 1) = 0$
$(\lambda + 1) + \lambda (\lambda + 1) (\lambda - 1) - 1(\lambda + 1) = 0$
$(\lambda + 1) + \lambda (\lambda + 1) (\lambda - 1) - 1(\lambda + 1) = 0$
Taking $(\lambda + 1)$ and simplifying further we get,
$\lambda (\lambda - 1)(\lambda + 1) = 0$
So we get
$\lambda$ $= 0,1, - 1$
So there are three values for which it is a non-trivial solution.

So, the correct answer is “Option D”.

Note: If set of the linear equations are given to be trivial then variables all are equal to zero but if it is a non-trivial solution then at least one must not be equal to zero.