Mathematica Solving system of equations Mathematica

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The discussion revolves around solving a system of equations represented in matrix form using Mathematica. The user encounters an issue where the command LinearSolve[] yields a particular solution represented as a 3D vector, despite the original matrix being 4D. The user seeks a method to obtain a complete solution that includes both the particular solution and the kernel's basis, expressed in a form that combines the particular solution with a linear combination of vectors from the null space. There is a suggestion that the determinant of the matrix is zero, indicating the presence of free variables. However, there is no clear solution provided for achieving the desired output in Mathematica, and the discussion highlights a lack of familiarity with the software's functions.
Mr Davis 97
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I am solving the following system of equations (in matrix form):

##\begin{pmatrix} 1 & -2 & -1 & 1 \\ 2 & -3 & 1 & 6 \\ 3 & -5 & 0 & 7 \\ 1 & 0 & 5 & 9 \end{pmatrix}##

I want to solve it using Mathematica, but when I use the command LinearSolve[], I only get back ##\begin{pmatrix} 9 \\ 4 \\ 0 \end{pmatrix}##, which is a particular solution. However, I am looking for how to get an output of something like ##\begin{pmatrix} 9 \\ 4 \\ 0 \end{pmatrix} + t_1\begin{pmatrix} -5 \\ -3 \\ 1 \end{pmatrix}##, where the linear combinations of the basis for the kernel is included. Is there a single command that does this? Or would I have to separately the NullSpace[] command?
 
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Hi Mr Davis:

I know next to nothing about Mathematica, but I would expect it to have a function you could invoke that would invert a matrix for you. Did you look for one to do that?

ADDED
Sorry, I misunderstood your question. Since you got a 3d vector as the solution from a 4d matrix, I am guessing the determinant is zero. I don't understand why the answer would be a 3d vector, so I am at a loss. I would expect, and I am guessing you are looking for, an answer of the form:
V1 + const x V2​
where V1 and V2 are 4d vectors. V1 is a solution to
M x V = V3,​
and V2 is a solution for
M x V = 0.​

I have nothing to suggest about how to approach this using Mathematica.

Regards,
Buzz
 
Last edited:

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