@synred Yes, certainly there's a simpler way to do it. The problem that I have Answering your Question is that there are several Maple packages that implement tensors, and there are several Maple packages that have commands named Multiply. I need to know which of each you're using. So, all I really need to see---and what I originally asked you for---are any with commands that you used and how you assigned a value to g5. And I need to know which packages and commands you're actually using, which may differ from which you think you're using. So I need to know the order that the with commands are given.

Regarding matrix multiplication, Does this work for you?

A:= <1, 2; 3, 4>; B:= <5, 6; 7, 8>;  #2x2 matrices
A . B;  #matrix-matrix multiplication
5 * A;  #scalar-matrix multiplication

@vs140580 No row in any of the three matrices that you just posted is an equivalence match (using your special forms of "equivalence") to any row in either of the other two matrices, which makes these not very good test cases. Please post some better matrices.

@synred I am asking you to upload *your* worksheet; I wasn't directing you to download some other worksheet.

Your Question shows an interaction between you and Maple, which I guess you took a screenshot of and put in your Question. You can save that interaction as a "worksheet". Just use the File -> Save dialog, pretty much like any other program. Then upload that saved worksheet here by using the green up arrow that's in the editor toolbar when you're posting here.

The portion of your work that you showed in your Question is not enough for me to be able to Answer it; I need to see the whole worksheet. 

Please upload a worksheet (as an attached file---use the green up arrow) that shows any with commands you used and how g5 was created.

@vs140580 I wrote a Maple program for this. I'd like to test it before posting it. Would you upload a Maple worksheet with some test matrices? My program is intended to work for any equivalence relation between rows for which there's an "easy" function which when passed a single row can identify the equivalence class that it belongs to. This includes the two row-equivalence relations already discussed in this thread.

It doesn't have a Taylor series for epsilon=0 because sqrt(epsilon) isn't differentiable at 0. You could use another expansion point though.

@pallav The error message gets truncated by being put in the title. Please transcribe the complete error message into the body of a message.

Also, since it worked for me in Maple 2022.1, I ask What version are you using?

@JAMET Don't make it seem as if the information that I gave you before was incorrect by saying that you still have errors. The truth is that you added the solve command, so you have new errors related to a new command that wasn't present for the question that I answered. My solution worked for the question as it was asked.

I'm having difficulty with combining assumptions (in assuming form) with the geometry package and then using solve. The variable a becomes localized in the equations, and solve sees two different variables a. It's much easier to not use geometry. Then assumptions aren't needed at all.

Since you have two equations and three variables (x, y, a), you need to tell solve to solve for {x, y} by giving {x, y} as the second argument.

restart:
b:= a*(1/2 + 1/6*sqrt(45 - 24*sqrt(3)))^2:
r:= b*sqrt(b)/(sqrt(a + b) + sqrt(a)):
E:= x^2/a^2 + y^2/b^2 = 1:
C3:= (x - (1+sqrt(3))*r)^2 + (y+r)^2 = r^2:
sol:= solve((simplify@numer@(lhs-rhs))~({E, C3}) union ({x, -y, a} >~ 0), {x, y}):
sol1:= simplify(sol) assuming a>0:
PP:= eval([x,y], sol1[]); 

@JAMET Use the command 

assume(a > 0);

somewhere before (the exact placement doesn't matter) the ellipse command.

@tomleslie My Reply was directed to the OP to answer their immediately preceding "Why...?" I wasn't criticizing your Answer, and I think that your use of fsolve was totally appropriate.

@JAMET PP was created with results from fsolve. Thus its coordinates are decimal numbers (aka "floats"). But ic1 is created with exact arithmetic. No floats would ever exactly satisfy its equation.

You have been editing this Question to the point of unanswerability. Stop. Add material rather than removing it.

First, this is a Question, not a Post. Don't put it in the Post section.

Second, your original Question was about a BVP with a delay. Maple doesn't handle those, hence the Reply that you received. Your removal of the original equations has made that Reply seem out of place. That's not fair. In other words, it's not fair to change your Question to such an extent that it makes someone else's response---which was originally relevant---seem irrelevant or like a non sequitur.

Third, you next Question removed the delay, but now you got the error "initial Newton iteration not converging." This is a common problem with BVPs, especially those of the boundary-layer variety, as yours is. That Question had enough information to be Answered; all the equations were there.

Finally, in its current state, your Question has no equations and no attached worksheet. As such, it's nearly impossible to Answer. Yes, I could direct you to the help page on "continuation parameters" for BVPs, which is usually the first step to solving this convergence error, but there are usually a few tricky steps to it.

@Steven_Huang At this point, your Question has no attached file, nor any inline code or equations.

@vs140580 Here is a summary of my understanding of your new "equality rule" (perhaps better called an "equivalence relation") that you want to implement, which I am recalling from memory of a single reading of the PDF that was  attached to your recently deleted Question. Please let me know if this is correct.

1. Given: Matrices A and B. If one has fewer columns than the other, then that one is padded on the right with columns of zeroes. The number of rows of the matrices is irrelevant (could be the same or different).
    
2. Given rows R1 of A and R2 of B, the rows are considered to be equivalent if and only if R1[2..-2] is a cyclic permutation of R2[2..-2]; the first and last entries of the rows are irrelevant.
   
   I am unsure about the following; please clarify: The permutation can be any cyclic permutation; it need not be a cyclic permutation of a single step. For example, R1 = [0, 1, 2, 3, 4, 0] and R2 = [0, 3, 4, 1, 2, 0] are equivalent.
    
3. As before, A and B are to be considered equivalent if and only if every row of A is equivalent to some row of B   AND   every row of B is equivalent to some row of A.
    
4. As before, you'd also like a set of pairs of integers (a,b) such that A[a] is equivalent to B[b]. (Note that when a matrix is given a single index, that's a row selection.) Thus, we are essentially constructing a bipartite graph matching the rows of A to those of B. The equivalence of A and B is the same as that graph having no isolated (i.e., degree-0) vertices.

Considering this and several of your previous Questions, it seems that you have numerous problems that fit the outline (1--4) given above with the only difference being point 2---the conditions that determine row equivalence. Is that correct?

@vs140580 If you wanted the row numbers, then you should've said so in the first place. I need to go to bed now. I'll work on the row numbers when I wake up. But, it'll be much much more complicated than the code I already gave.