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Modulus of a number cannot be negative. A simple idea that somehow escapes our mind every now and then.

## Unique Solutions

a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + c3z = d3.

Which of the following statements if true would imply that the above system of equations does not have a unique solution?
i. ${\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \neq \frac{d_{1}}{d_{2}}}$
ii. ${\frac{ a_{1} }{ a_{2} }= \frac{ a_{2} }{ a_{3} }}$ ; ${\frac{ b_{1} }{ b_{2} }= \frac{ b_{2} }{ b_{3} }}$
iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers

Statement i.

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## Detailed Solution

If we have three independent equations, we will have a unique solution. In other words, we will not have unique solutions if:
The equations are inconsistent or
Two equations can be combined to give the third

Now, let us move to the statements.

i. ${\frac{ a_{1} }{ a_{2} }= \frac{ b_{1} }{ a_{2} } = \frac{ c_{1} }{ c_{2} } \neq \frac{ d_{1} }{ d_{2} }}$

This tells us that the first two equations cannot hold good at the same time.

x + y + z = 3;
2x + 2y + 2z = 5.
Either the first or the second can hold good. Both cannot hold good at the same time. So, this will definitely not have any solution.

ii. ${\frac{ a_{1} }{ a_{2} }= \frac{ a_{2} }{ a_{3} }}$ and ${\frac{ b_{1} }{ b_{2} }= \frac{ b_{2} }{ b_{3} }}$

a1, a2, a3 are in GP, b1, b2, b3. This does not prevent the system from having a unique solution.

For instance, if we have

x + 9y + 5z = 11
2x + 3y – 6z = 17
4x + y – 3z = 15

This could very well have a unique solution.

iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers
This gives us practically nothing. This system of equations can definitely have a unique solution.

So, only Statement i tells us that a unique solution is impossible.

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