Why Does a Linear System Have a Single Solution?

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why lenear system has a single solution?
 
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In general, a linear system of equations can have no solution, a single solution, or an infinite number of solutions. Do you have a question about a particular linear system?
 
it for electrical circuit theory

for thevenin theorem

i have a pi system which has independent sources

and my prof said

"we have a lenear system so the is a single solution"

i can't understand why?
 
I don't know what a "pi system" is. In any case, unless you know something about the linear system, you can't tell how many solutions it has.

Here are three very simple examples:

1. No solutions
x + 2y = 3
x + 2y = 4

2. Exactly one solution
x + y = 3
x - y = 1
(solution is x = 2, y = 1)

3. Infinite number of solutions
x + y = 3
2x + 2y = 6
 
Note Mark44's options: one solution, no solution, infinite number of solutions.

If you know the problem has a non-zero finite number of solutions then you can say there is a single solution- as opposed to, say, a quadratic equation in which there can be two solutions.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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