Question

The system of equations. $ar+y+z=a-1$, $x+ay+z=a-1$ and $x+y+az=a-1$ doesn't have unique solution if a=

• A. $1$ or $-2$
• B. $3$
• C. $2$
• D. $-1$

Answer 1

The correct option is A $1$ or $-2$

The condition for system of equations, to not have unique solution, is

$\Delta =0$

Given,

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

$det\left(\begin{array}{ccc}a& 1& 1\\ 1& a& 1\\ 1& 1& a\end{array}\right)$

$=adet\left(\begin{array}{cc}a& 1\\ 1& a\end{array}\right)-1\cdot det\left(\begin{array}{cc}1& 1\\ 1& a\end{array}\right)+1\cdot det\left(\begin{array}{cc}1& a\\ 1& 1\end{array}\right)$

$=a(a^2-1)-1\cdot (a-1)+1\cdot (1-a)$

$=a^3-3a+2$

$\Rightarrow a^3-3a+2=0$

${(a-1)}^2(a+2)=0$

$\therefore a=1,a=-2$