# Solving linear equation using maple [closed]

Use Maple to find the general solution to the following system of linear equations: $$\begin{eqnarray} 2x1 + x2 - x3 + 3x4 = 2\\ x1 + 2x2 - x4 = -1\\ 3x1 + 2x2 - 2x3 + x4 = 1\\ \end{eqnarray}$$

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## closed as off-topic by Jonas Meyer, pizza, Sujaan Kunalan, user91500, MPOMar 24 at 6:22

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – pizza, Sujaan Kunalan, user91500, MPO
If this question can be reworded to fit the rules in the help center, please edit the question.

-1: "This question does not show any research effort; it is unclear or not useful" –  Alex Becker Mar 2 '12 at 13:31
Perhaps phrasing this as a question (rather than a demand) will make people more willing to help. –  Holdsworth88 Mar 2 '12 at 13:44

With rank of 3, this system of three equations in four variables has 1 free parameter in the solution.

with(LinearAlgebra):

eqs:={x2 = 3*x4-2, x1 = -5*x4+3, x3 = -4*x4+2}:

A,b:=GenerateMatrix(eqs,[x1,x2,x3,x4]):

Rank(A);

3

X:=LinearSolve(A,b,'free'=t):

Equate([x1,x2,x3,x4],X);

[x1 = 3 - 5 t[4], x2 = -2 + 3 t[4], x3 = 2 - 4 t[4], x4 = t[4]]

solve(eqs);

{x1 = -5 x4 + 3, x2 = 3 x4 - 2, x3 = -4 x4 + 2, x4 = x4}


Using LinearSolve, we can choose how such free parameters get named. Using solve instead, the degree or freedom is illustrated by the x4=x4 term in the solution. This all means that there are infinitely many solutions, depending on whatever value is taken for the free parameter.

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  sys:={2*a+b-c+3*d=2,a+2*b-d=-1,3*a+2*b-2*c=1};
solve(sys);

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Please note that you could write codes as I had done in the edit. To access the edit, you may want to click on the time stamp. –  user21436 Mar 3 '12 at 7:17