A) \[R-\{1\}\]
B) \[R-\{-1\}\]
C) \[\{1,-1\}\]
D) \[\{1,0,-1\}\]
Correct Answer: B
Solution :
Given system of equations is homogeneous which is \[x+ay=0\] \[y+az=0\] \[z+ax=0\] It can be written in matrix form as \[A=\left( \begin{matrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \\ \end{matrix} \right)\] Now\[|A|=[1-a(-{{a}^{2}})]=1+{{a}^{3}}\ne 0\] So, system has only trivial solution. Now, |A| = 0 only when \[a=-1\] So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solution. Hence set of all real values of ?a' is \[R-\{-1\}.\]
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