Question

The system of equations , $ax+y+z=a-1$ , $x+ay+z=a-1$, $x+y+az=a-1$has no solution, if a is

• A. either $-2or1$
• B. $-2$
• C. $1$
• D. $not-2$

Answer 1

Answer: (A)

either $-2or1$

Determinant of Coefficient Matrix , $\left[\begin{array}{ccc}a& 1& 1\\ 1& a& 1\\ 1& 1& a\end{array}\right]=0$

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

$a^3-a-a+1+1-a=0$

$a^3-3a+2=0\Rightarrow$ If $a=1$

$a=1|\begin{array}{cccc}1& 0& -3& 2\\ 0& 1& 1& -2\end{array}$

|$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$\begin{array}{cccc}1& 1& -2& 0\end{array}$

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

$a(a+2)-1(a+2)=0$

$a=-2,1$

$a$ is either $-2$ (or) $1$