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(A) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(

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a singleton set

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(A) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.The determinant $D$ is given by:$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & a & 1 \ a & b & 1 \end{vmatrix} = 1(a - b) - 1(1 - a) + 1(b - a^2) = a - b - 1 + a + b - a^2 = -a^2 + 2a - 1 = -(a - 1)^2$.For the system to have no solution or infinitely many solutions,we must have $D = 0$,which implies $-(a - 1)^2 = 0$,so $a = 1$.Substituting $a = 1$ into the system:$x + y + z = 1$$x + y + z = 1$$x + by + z = 0$For the system to have no solution,the third equation must be inconsistent with the first two. Since the first two equations represent the same plane $x + y + z = 1$,the third equation $x + by + z = 0$ must represent a plane parallel to the first one but not identical to it.Comparing $x + y + z = 1$ and $x + by + z = 0$,we see that for the planes to be parallel,the coefficients of $x, y, z$ must be proportional. Thus,$1/1 = b/1 = 1/1$,which implies $b = 1$.However,if $b = 1$,the third equation becomes $x + y + z = 0$,which is parallel to $x + y + z = 1$ but distinct (since $0 
eq 1$). Thus,the system has no solution when $b = 1$.Therefore,$S = \{1\}$,which is a singleton set.

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

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