4) now i want to solve this like that: A*B=C
my idea was:
but it doesnt work...
solve doesn't work because it expects lists as its arguments. However, there are many ways to solve a system of linear equations:The simplest way is to multiply the inverse of A with C:
A^-1 * C
If you really want to use solve, you must convert your system to a list of equations first. Typing
will create a list of equations. Now, normally you could solve the system by typing
but solve returns "Wrong argument type", since their arguments must be lists; variables equal to lists are not accepted (this is a known serious limitation). So if you want to use solve, you must substitute syst and matToList(X,1) by their equivalent lists, as printed in the output. As you can see, it is possible to use the function solve for systems of linear equations, but it is not convenient at all.
you can use a funtion called rref that solves the ecuation i thing it uses the gaus jordan methoth because it returns the answer and the indentiti matrix, you cant see it at the classpad manual is very easy to use...
Well, it is possible to solve the system this way, although it is not the simplest way to proceed, plus I don't think that it helps our friend too much. Nevertheless, since you mentioned it, and you don't give details, here is how to use rref to solve a system: Augment the matrix A by a column equal to C, compute the reduced row echelon form, then extract the last column, which is the solution. All this can be done in one expression:
The main disadvantage is that the number "4" in the above expression is valid for a 3 x 3 system only. In general, this number should be replaced by n+1 for a n x n system. You can avoid this by using the alternate (and more complex) expressions
You can, of course, avoid to write two expressions:
but this is more time-consuming, since rref(augment(A,C)) is computed twice. As you can see, using rref it is not
the simplest way to proceed.
I can also post other ways to solve the system, e.g., using LU or QR decompositions, but is there any sense to proceed this way?General conclusion: If you need to solve a system of linear equations, multiply the inverse of the matrix A with the constants-matrix C. It's definitely simplest than any alternative.