Q. Consider the system of equations $ax+y=b,bx+y=2a\&ax+by=3ab$ , where $a,b\in \left\{\right.0,1,2\left.\right\}$ . The number of ordered pairs $\left(\right.a,b\left.\right)$ for which the system of equations is consistent, is equal to

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$0$

$1$

$2$

$3$

Solution:

For consistency $\begin{vmatrix} a & 1 & b \\ b & 1 & 2a \\ a & b & 3ab \end{vmatrix}=0$
$a^{2}b-3ab^{2}+2a^{2}+b^{3}-ab=0$
$b^{3}-3ab^{2}+\left(a^{2} - a\right)b+2a^{2}=0$
$b=0:a=0$
$b=1;1-3a+a^{2}-a+2a^{2}=0$
$3a^{2}-4a+1=0$
$\left(\right.3a-1\left.\right)\left(\right.a-1\left.\right)=0\Rightarrow a=1\left(\right.$ reject $\left.$ or $a=\frac{1}{3}$ $b=2:8-12a+2a^{2}-2a+2a^{2}=0$
$4a^{2}-14a+8=0$
$2a^{2}-7a+4=0$
$\Rightarrow $ not integral solution only one ordered pair $\left(\right.0,0\left.\right)$

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