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Consider the system of equations:

 x+ay=0, \; y+az=0 and z+ax=0. Then the set of all real values of 'a' for which the system has a unique solution is :

Option: 1

     R-\left \{1 \}             


Option: 2

R-\left \{-1 \}


Option: 3

\{1,\; -1 \}


Option: 4

\{1,0,-1 \}


Answers (1)

best_answer

Given system of equation is homogeneous, which is 

\\x+a y=0 \\ y+a z=0 \\ z+a x=0

\\A=\begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} \\|\mathrm{A}|=\left[1-a\left(-a^{2}\right)\right]=1+a^{3} \neq 0

So, the system has only trivial solution.

Now, |\mathrm{A}|=0$ only when $a=-1

Posted by

Sayak

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