## Solve Ax=b in Maxima, part 2

In a previous post, I included my little coding project to implement a general backsolve() function to use with the built-in maxima matrix function echelon(), producing an easy-to-call matrix solver matsolve(A,b).  The result is meant to solve a general matrix vector equation $Ax=b$ , including cases when $A$ is non-square and/or non-invertible.

Here’s a quicker approach — convert the matrix into an explicit system of equations using a vector of dummy variables, feed the result into the built-in Maxima function linsolve(), and then extract the right hand sides of the resulting solutions and put them into a column vector.

The two methods often behave identically, but here’s an example that breaks the linsolve() method, where the backsolve() method gives a correct solution:

*Note, I’ve found that the symbol rhs is a very popular thing for users to call their problem-specific vectors or functions.  Maxima’s  “all symbols are global” bug/feature generally wouldn’t cause a problem with a function call to rhs(), but the function map(rhs, list of equations)  ignores that rhs() is a function and uses user-defined rhs.  For that reason I protect that name in the block declarations so that rhs() works as expected in the map() line at the bottom.  I think I could have done the same thing with a quote: map(‘rhs, list of equations).

matsolve2(A,b):=block(
[rhs,inp,sol,Ax,m,n,vars],
[m,n]:[length(A),length(transpose(A))],
vars:makelist(xx[i],i,1,n,1),
Ax:A.vars,
inp:makelist(part(Ax,i,1)=b[i],i,1,n,1),
sol:linsolve(inp,vars),
expand(transpose(matrix(map(rhs,sol))))
);

## Solving the matrix vector equation Ax=b in Maxima

*Upadate:  I’ve implemented a Maxima matrix-vector equation solver with simpler Maxima-specific algorithm in a later post.  That method is based on the built-in function linsolve().  In that post I show an example that breaks linsolve() but that is handled correctly by the backsolve() method.

Is there really not a solver in Maxima that takes matrix A and vector b and returns the solution of $Ax=b$ ?  Of course we could do invert(A).b, but that ignores consistent systems where $A$ isn’t invertible…or even isn’t square.

Here’s a little function matsolve(A,b)  that solves $Ax=b$ for general $A$ using the built-in Gaussian Elimination routine echelon(), with the addition of a homemade backsolve() function.  The function in turn relies on a little pivot column detector pivot() and my matrix dimension utility matsize(). This should include the possibilities of non-square $A$, non-invertible $A$, and treats the case of non-unique solutions in a more or less systematic way.

matsolve(A,b):=block(
[AugU],
backsolve(AugU)
);

backsolve(augU):=block(
[i,j,m,n,b,x,klist,k,np,nosoln:false],
[m,n]:matsize(augU),
b:col(augU,n),
klist:makelist(concat('%k,i),i,1,n-1),
k:0,
x:transpose(matrix(klist)),
for i:m thru 1 step -1 do (
np:pivot(row(augU,i)),
if is(equal(np,n)) then
(nosoln:true,return())
else if not(is(equal(np,0))) then
(x[np]:b[i],
for j:np+1 thru n-1 do
x[np]:x[np]-augU[i,j]*x[j])
),
if nosoln then
return([])
else
return(expand(x))
)\$

matsize(A):=[length(A),length(transpose(A))]\$

pivot(rr):=block([i,rlen],
p:0,
rlen:length(transpose(rr)),
for i:1 thru rlen do(
if is(equal(part(rr,1,i),1)) then (p:i,return())),
return(p)
)\$