Consider the system of equations : $x+a y=0, y+a z=0$ and $z+a x=0$. Then the set of all real values of '…
Consider the system of equations :
$x+a y=0, y+a z=0$ and $z+a x=0$. Then the set of all real values of ' $a$ ' for which the system has a unique solution is:
$\mathrm{R}-\{1\}$
$\mathrm{R}-\{-1\}$
$\{1,-1\}$
$\{1,0,-1\}$
Solution
Given system of equations is homogeneous which is
$
\begin{aligned}
&x+a y=0 \\
&y+a z=0 \\
&z+a x=0
\end{aligned}
$
It can be written in matrix form as
$
\mathrm{A}=\left(\begin{array}{lll}
1 & a & 0 \\
0 & 1 & a \\
a & 0 & 1
\end{array}\right)
$
Now, $|\mathrm{A}|=\left[1-a\left(-a^2\right)\right]=1+a^3 \neq 0$
So, system has only trivial solution.
Now, $|\mathrm{A}|=0$ 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
$
\mathrm{R}-\{-1\}
$