$ \Rightarrow 1\left[ {a - b} \right]\, - 1\left[ {1 - a} \right] + 1\left[ {b - {a^2}} \right] = 0 \Rightarrow {\left( {a - 1} \right)^2} = 0$

$ \Rightarrow a = 1$

For $ \Rightarrow a = 1$, First two equation are identical ie. $x+y+z=1$

To have no solution with $x+by+z=0$

$b=1$

So $b = \left\{ 1 \right\} \Rightarrow $ It is singleton set.