System of Equations with $\mu$ .

Algebra Level 4

$\large\begin{cases} x + \mu y - z =0\\ \mu x \ - y - z =0\\ x + y - \mu z=0 \end{cases}$

The system of linear equations above has a non-trivial solution for:

$\quad$ a) Exactly three values of $\mu$ .

$\quad$ b) Infinitely many values of $\mu$ .

$\quad$ c) Exactly one value of $\mu$ .

$\quad$ d) Exactly two values of $\mu$ .

Problem courtesy: Online Indian Exam

b

d

c

Not enough information.

a

1 solution

Hana Wehbi
Jun 12, 2018

Clearly for a trivial solution, we have $x=0, y=0 \text{ and } z=0$ since the system is homogeneous.

For a non-trivial solution and this type of solution only exists if the determinate is 0. Therefore, to consider the possible values of $\mu$ incase of a nontrivial solution, we have to set the determinate equals to 0 and solve for $\mu$ as follows:

$\begin{vmatrix} 1 & \mu & -1\\ \mu & -1 & -1\\ 1 & 1& -\mu \\ \end{vmatrix}=0 \implies \mu^3 - \mu = 0 \implies \mu(\mu -1)(\mu+1)=0 \implies \mu=0,1 \text{ and } -1$

Which indicates three values for $\mu$ .

You should make it clear that "A non-trivial solution exists if and only if the determinant is 0". Otherwise, we do not know if
1) all of these 3 values work,
2) no other values could work.

In part, having one-way implication signs also suggest that you only have a superset of possible values, but still need to check that all of these are valid solutions to the problem.

Calvin Lin Staff - 2 years, 12 months ago

Ok, l will add few things to my solution. Thank you for pointing it out.

Hana Wehbi - 2 years, 12 months ago

System of Equations with μ \mu μ .

1 solution

0 pending reports

System of Equations with $\mu$ .