Question

Find the value of ‘$c$’ for which the following equations have non-trivial solutions: $cx-y-z=0,-cx+y-cz=0,x+y-cz=0$.

• A. $\frac{1}{2}$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is B

$-1$

Finding the value of C :

Given equation, $cx-y-z=0,-cx+y-cz=0,x+y-cz=0$

Convert those three equation in determinant form:

For non-trivial solutions of the given three equations:

$\left|\begin{array}{ccc}c& -1& -1\\ -c& 1& -c\\ 1& 1& -c\end{array}\right|=0$

$\Rightarrow$$c(-c+c)+1(c^2+c)-1(-c-1)=0$

$\Rightarrow$ $c^2+c+c+1=0$

$\Rightarrow$ ${(c+1)}^2=0$

$\Rightarrow$ $(c+1)=0$

$\mathbf{\therefore }\mathit{c}\mathbf{=}\mathbf{-}\mathbf{1}$

Hence, Option (B) is the correct answer.