Question

Consider the system of equations: $x+ay=0$, $y+az=0$ and $z+ax=0$. Then the set of all real values of ${}^{\prime }{a}^{\prime }$ for which the system has a unique solution is

• A. $\{1,-1\}$
• B. $R-\{-1\}$
• C. $\{1,0,-1\}$
• D. $R-\left\{1\right\}$

Answer 1

Answer: (B)

$R-\{-1\}$

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

$A=\left(\begin{array}{ccc}1& a& 0\\ 0& 1& a\\ a& 0& 1\end{array}\right)$

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 ${}^{\prime }{a}^{\prime }$ is $R-\{-1\}$