Engineer98
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Does this System have infinite or finite solutions?
Does that depend on i?
Number of solutions for the following Equations System: x_i A_i + y_i B_i = C_i, ...
- Thread starter Engineer98
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Does this System have infinite or finite solutions?
Does that depend on i?
Dr.Peterson
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For n = 1, 2, 3, how many variables do you have? How many equations are there?
That will suggest an answer.
There is no general answer to your question, but it has nothing to do with i or n.
In general, a system of equations will not have any solution if one or more equations are inconsistent with one or more different equations in the system or may have an infinite number of solutions if two or more equations are not independent. With a system of linear equations in n unknowns, there generally is a unique solution given n independent equations. Independence of specific linear equations can be tested easily: if any equation is an exact multiple of another equation, those equations are not independent. But independence cannot be determined with respect to an unspecified system such as yours. Nor can inconsistency. What is particularly worrying about your question is that you have n + 2 equations. At least two must either be inconsistent or dependent. If you have inconsistency, the problem is insoluble. If you have dependence, there may be an infinite number of solutions. But with n + 2 linear equations, you may be able to eliminate the dependent equations and get a unique answer; that question is, as the lawyers say, fact-specific rather than general.
If you are modelling a physical situation that you know has a unique physical solution, you need to figure out which two equations to eliminate.
Inconsistent system:
[math]x = y + 1 \text { and } x = y - 1 \implies \\ x - x = y + 1 - (y - 1) \implies \\ 0 = 2.[/math]
Dependent system:
[math]8x + 4y = 100 \text { and } 40x + 20y = 500 \implies \\ y = 25 - 2x, \text { a solution set with infinite members}.[/math]
In general, a system of equations will not have any solution if one or more equations are inconsistent with one or more different equations in the system or may have an infinite number of solutions if two or more equations are not independent. With a system of linear equations in n unknowns, there generally is a unique solution given n independent equations. Independence of specific linear equations can be tested easily: if any equation is an exact multiple of another equation, those equations are not independent. But independence cannot be determined with respect to an unspecified system such as yours. Nor can inconsistency. What is particularly worrying about your question is that you have n + 2 equations. At least two must either be inconsistent or dependent. If you have inconsistency, the problem is insoluble. If you have dependence, there may be an infinite number of solutions. But with n + 2 linear equations, you may be able to eliminate the dependent equations and get a unique answer; that question is, as the lawyers say, fact-specific rather than general.
If you are modelling a physical situation that you know has a unique physical solution, you need to figure out which two equations to eliminate.
Inconsistent system:
[math]x = y + 1 \text { and } x = y - 1 \implies \\ x - x = y + 1 - (y - 1) \implies \\ 0 = 2.[/math]
Dependent system:
[math]8x + 4y = 100 \text { and } 40x + 20y = 500 \implies \\ y = 25 - 2x, \text { a solution set with infinite members}.[/math]
Last edited:
Dr.Peterson
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The fact I'm waiting for the OP to see is that there are actually 2n unknowns (n x's and n y's) and n+2 equations, so the n=1, 2, and 3 cases are all different. Writing those cases out, and perhaps considering actual values for the constants, should be very enlightening.
Oh, right. I'll delete my post.