if the system of equations
`(a-t)x+by +cz=0`
`bx+(c-t) y+az=0`
`cx+ay+(b-t)z=0`
has non-trivial solutions then product of all possible values of t is

Options

A. `|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|`
B. `a+b+c`
C. `a^(2)+b^(2)+c^(2)`
D. `1`

Answer

A

Explanation

The given system of equations will have a non-trivial solutions if the determinant of coefficient is 0.
`Delta= |{:(a-t,,b,,c),(b,,c-t,,a),(c,,a,,b-t):}|=0`
`Delta=0` is a cubic equation in t, so it has 3 solutions say `t_(1), t_(2)" and " t_(3)`
Let `Delta =p_(0)t_(3)+p_(1)t^(2) +p_(2)t+p_(3)`
Clearly ,Po= coeff . of `t^(3)` which is equal to -1 , so
`t_(1) t_(2)t_(3) =-(P_(3))/((-1))=P_(3)`
= constant term in the expansion of `Delta i.e, Delta _((t=0))`
hence `t_(1)t_(2)t_(3)= |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|`