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(C) For a system of linear equations $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1

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$2$

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(C) For a system of linear equations $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$.Given equations are $a^2 x + (2 - a) y = 4 + a^2$ and $a x + (2 a - 1) y = a^5 - 2$.Applying the condition: $\frac{a^2}{a} = \frac{2 - a}{2a - 1} 
eq \frac{4 + a^2}{a^5 - 2}$.From $\frac{a^2}{a} = \frac{2 - a}{2a - 1}$ (assuming $a 
eq 0$): $a = \frac{2 - a}{2a - 1} \implies 2a^2 - a = 2 - a \implies 2a^2 = 2 \implies a^2 = 1 \implies a = 1$ or $a = -1$.Case $1$: If $a = 1$,the ratio is $\frac{1}{1} = \frac{1}{1} 
eq \frac{5}{-1}$,which holds true.Case $2$: If $a = -1$,the ratio is $\frac{1}{-1} = \frac{3}{-3} 
eq \frac{5}{-3}$,which holds true.If $a = 0$,the equations become $2y = 4$ and $-y = -2$,which gives $y = 2$,so there is a solution. Thus $a=0$ is not a solution.Therefore,there are $2$ values of $a$ for which the system has no solution.

The number of values of $a$ for which the system of equations $a^2 x + (2 - a) y = 4 + a^2$ and $a x + (2 a - 1) y = a^5 - 2$ possesses no solution is:

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