Question

If S is the set of distinct values of b for which the folowing system of linear equations
$x+y+z=1$,
$x+ay+z=1$
and $ax+by+z=0$ has no solution, then S is

• A. an infinite set
• B. a finite set containing two or more elements
• C. singleton set
• D. a empty set

Answer 1

The correct option is C singleton set

$\because \Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ a& b& 1\end{array}\right|=1(a-b)-1(1-a)+1(b-a^2)=-(a-1{)}^2{\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ 0& b& 1\end{array}\right|=1(a-b)-1(1)+1(b)=-(a-1){\Delta }_2=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ a& 0& 1\end{array}\right|=1(1)-1(1-a)+1(1-a)=0$
$\text{and}{\Delta }_3=\left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ a& b& 0\end{array}\right|=1(-b)-1(-a)+1(b-a^2)=-a(a-1)$
For $a=1$
$\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$
$x+y+z=0$,
$x+y+z=1$
and $x+y+z=0$
i.e. no solution ($\because$ RHS is not equal)
Hence, for no solution b = 1 only