Like this:

restart:
oldGAMMA:= eval(GAMMA):
unprotect(GAMMA):
GAMMA:= overload([
    proc(x::And(integer, Not(0)), y::And(algebraic, Not(complexcons)), $)
    option overload;
        'procname'(x,y)
    end proc,
    oldGAMMA
]):
protect(GAMMA, oldGAMMA)
:
#Test:
GAMMA(-1,x), oldGAMMA(-1,x), GAMMA(3);

Like this (I've divided all the quantities by 1000 to reduce visual clutter):
 

Download LPTransNetw.mw

@Earl (Perhaps that title should be "A get-Maple-to-do-more-algebra approach.") I should've shown that simplification of W. I think that that would've avoided that version difference of implicitplot. And it'll make your attempt to solve W for y work better. Indeed, by solving for x and y separately and plotting both, you can get a perfect plot (shown below).

We simplify W by using the fact that it's a rational function of x and y implicitly equated to 0. Thus,

1. Its denominator doesn't matter and can be discarded. This is the purpose of numer below.
2. The "content" (common factors of all numerator terms) can be discarded. This is the purpose of op&<2 in 
   op~&<1 @ op&<2 @ evala@AFactors below.  (My &< is defined below.) That @-expression is equivalent to
   e-> op~(1, evala(AFactors(e))[2]).
3. Exponents on factors of the numerator can be discarded. This is the purpose of 
   op~&<1. 

restart
:
#abbreviations for things that I commonly use in @-expressions:
`&<`:= curry: `&>`:= rcurry:   
:
WattEq:= r^2 - (b^2 - (a*sin(theta) + sqrt(c^2 - a^2*cos(theta)^2))^2);


#Don't use decimal values for the parameters.
params:= (`=`~ &< [a,b,c])~([[31,11,30]/10, [1,sqrt(2),1], [21,22,6]/10]);


rng:= -b..b:
V:= [x,y]:
abc:= e-> eval(e, params[Params]):
 
Params:= 1: #which parameter values to use

W:= (op~&<1 @ op&<2 @ evala@AFactors @ numer @ evala@Norm @ eval[recurse])(
    abc(WattEq),
    #polar to cartesian: 
    [r= sqrt(x^2+y^2), (cos,sin)(theta)=~ V[]/~r]
):
if nops(W)<>1 then WARNING("multiple factors") fi:
W:= mul(W);


plots:-implicitplot(abc([W, V[]=~ rng])[], scaling= constrained);

For the purely functional/algebraic plot, we continue with the same W, and solve it for x and y separately. This'll lead to much duplication of the plotted points, but that doesn't matter; the gaps left by each will be filled by the other. The solutions for x are plotted in parametric form (e.g., plot([f(y), y, y= a..b])) so that the axes are switched. The solutions are substituted into a procedure template that does hardware-float evaluation. An attempt is made to distinguish these two cases:

1. If a value e has an imaginary part of very small magnitude, it's assumed that that is due solely to rounding error and Re(e) is returned;
2. If there's an imaginary part of larger magnitude, it's assumed that the value is truly non-real and undefined is returned.

The plot command doesn't care if some points are included that have undefined as one of their coordinates. It simply skips them. However, if a function has an undefined coordinate for all of its points, plot will issue a warning. To avoid these warnings, I remove those functions that always produce non-real values. There are 12 functions to start (as can be seen from the degree of W in x and y), and this filtering removes 4 of them.

So, here's the code that does all that: 

Digits:= 15: eps:= 1e-13: Ev:= evalhf: #All 3 may need to be adjusted.

#some testing values to decide real-or-complex issue; 99 is arbitrary:
T:= abc(b)/99 *~ {$1..99}:
 
(Px,Py):= (remove &< (w-> w~(T)={undefined}))~(
    (((w-> subs~&<(_P=~ w)) @ subs&<(V=~ z) @ [solve])~(W, V))( 
        z-> local e:= Ev(_P); `if`(abs(Im(e)) < eps, Re(e), undefined)
    )
)[]:
plot([`[]`~(Px, z->z, abc(rng))[], Py[]], abc(rng), scaling= constrained);

the

The above takes 0.516 seconds on my computer, and, because using evalhf makes the filtering and plotting nearly instantaneous, the vast majority of that time is spent in solve.

There's a chance (version-dependent I guess) that the solve will return RootOf expressions. In that case, add the explicit option. The polynomial is bicubic in both x and y, so it definitely can be explicitly solved.

There's a chance that some spurious horizontal lines will appear on the plot. These can be corrected by adjusting Digits, eps, or both. I haven't investigated the origins of these spurious lines. There's a remote chance, I guess, that you'll need to change evalhf to evalf to get higher precision, although I haven't yet seen a case where that's actually needed.

First, here are some things that you're doing wrong in your code snippet:

1. By row, I assume that you mean Row from LinearAlgebra. Note that its name is capitalized.
2. When the left operand of := is a function call (such as your row(out, k)), it is not evaluated (i.e., the function is not called) before the assignment^[*1]. Rather, these assignments are stored in something called the remember table of row. (At the moment, it's not useful to explain remember tables further; but I'll be happy to when it becomes actually useful.)
3. To do an assignment to a left operand that needs to be evaluated first, use assign, as in assign(f(x)= y).
4. I assume that out is a matrix, but it would need to be already created, and have the correct dimensions for your idea to have any chance of working. (And there are several other reasons why this won't work, but they're too complicated to explain right now. Anyway, there are several other easy ways to do what you want.)

Second, just for the sake of showing you some briefer syntax, here's a slight variation on Joe's Answer:

#shortcut "hand-typed" matrix constructor syntax:
M:= <1, 2, "A"; 2, 3, "B"; 3, 4, "A">;

M[[seq](`if`(M[i,3]="A", i, NULL), i= [rtable_dims](M)[1])];

Third, there's a matrix-based superstructure called a DataFrame that's perfect for fancy row-selection operations such as I suspect you have in mind. A DataFrame is just an Object whose underlying structure is an ordinary Matrix with many OOP-style fancy indexing methods overloaded onto the regular square-bracket indexing operator.

In the code below, M is the matrix from above, and columns is a list of names for the columns. There's nothing special about the names that I chose ([1, 2, key]).

DM:= DataFrame(M, columns= [1, 2, key]):
DM[DM[key]=~ "A"];

Footnote [*1]: There's only one exception to this that I'm aware of: When function cat is the left operand of :=, then it IS evaluated before the assignment is made, as in

cat(x, 3):= 4;

which assigns 4 to x3.

 

Like this:

W:= `[]`~(P, Q);

An operator is called a binary infix operator if it can take 2 arguments and can be placed between (rather than in front of) them. So, most familar arithmetic operators (such as +, -, *, /, =, <, ^, and others) are binary infix operators. The way that you wrote your Question suggests that you want to use oper in infix form. This is easy in Maple, and it's one of my favorite features of Maple's syntax. The operator's name should begin with &, followed by letters or a wide variety of punctuation marks.

AnyOp:= proc(`&op`, x, y) x &op y end proc:
AnyOp(`*`, 7, 3), AnyOp(`+`, 7, 3); 
                             21, 10

When the operator is not being used in infix form, it must be in back quotes (notice `&op` immediately after proc). This is true for all infix operators, not just those beginning with & (notice `*` and `+` above). On the other hand, when they are used in infix form, they must NOT have quotes.

In the examples above, the operation that is passed to `&op` need not be infix as long as it can accept two arguments (see iquo and irem below):

AnyOp:= (`&?`, x, y)-> x &? y:
AnyOp~([`*`, `+`, `-`, `/`, `mod`, iquo, irem, `^`], 7, 3);
                  [21, 10, 4, 7/3, 1, 2, 1, 343]

Since mod is one of Maple's infix operators, even though it's just alphabetic characters, it needs the quotes when it's passed.

The plot can be made perfectly (no gaps, no jaggedness) and easily (no fancy options or extra points or precision needed) by these steps:

1. Express your Watt equation as an expression implicitly equated to 0.
2. Convert the expression from polar coordinates to Cartesian (x,y). (There's a command for that, but rather than looking up its syntax, I just used the usual well-known substitutions.)
3. Rationalize the resulting algebraic function with evala@Norm. (The resulting rational function can be simplified significantly, which might be useful if you're going to do further algebra on it; but, it plots with no trouble without simplification.)
4. Put in your numeric parameters.
5. Plot with implicitplot.

restart:
WattEq:= r^2 - (b^2 - (a*sin(theta) + sqrt(c^2 - a^2*cos(theta)^2))^2);
params:= [a= 3.1, b= 1.1, c= 3.]:  rng:= -1.2..1.2:
W:= (evala @ Norm @ eval[recurse])(
    WattEq,
    #polar to cartesian: 
    [r= sqrt(x^2+y^2), (cos,sin)(theta)=~ (x,y)/~r]
);
  

plots:-implicitplot(eval(W, params), (x,y)=~ rng, scaling= constrained);

The most-basic way to remove specific cases of inertness is to change a name beginning with % to the equivalent name without a %. Thus, this handles your case:

Set:= eval({op}(InertForm:-Parse(Str)), %Vector= Vector);
InertForm:-ToMathML(`%+`(Set[]));
 

Here are two procedures: The first detects whether its argument is an overload, and the second deconstructs an overloaded argument into its original procedures.

IsOverload:= (P::procedure)-> 
    evalb(indets(ToInert(eval(P)), specfunc(_Inert_OVERLOADLIST)) <> {})
:
UnOverload:= (P::And(procedure, satisfies(IsOverload)))->
    indets(
        subsindets(
            ToInert(eval(P)), 
            specfunc(_Inert_OVERLOADLIST), 
            %overload@[FromInert]@op
        ),
        specfunc(%overload)
    )[-1]
:

#Example usage:

P:= overload([
    proc(n::And(posint, even)) option overload; n/2 end proc,
    proc(n::And(posint, odd)) option overload; 3*n+1 end proc,
    proc(n) 'procname'(args) end proc
]);
P := overload(
  [
  proc (n::(And(posint, even))) option overload; (1/2)*n end proc, 
  proc (n::(And(posint, odd))) option overload; 3*n+1 end proc, 
  proc (n) ('procname')(args) end proc])

IsOverload~([P, cos]);
                         [true, false]
UnOverload(P);
%overload(
  [
  proc (n::(And(posint, even))) option overload; (1/2)*n end proc, 
  proc (n::(And(posint, odd))) option overload; 3*n+1 end proc, 
  proc (n) ('procname')(args) end proc])

Replace * with %*:

P:= a %* b %* c %* d %* e;

In your 7-year-old version of Maple, you may need to use prefix form:

P:= `%*`(a, b, c, d, e);

Here's another way to do the unit conversion on a vector or matrix that I suspect is closer to what you were originally trying. Enter your vector, highlight and right click, do the unit conversion as you originally tried. When that's done, the vector will appear unchanged, as you initially reported. Now, give the command

rtable_eval(%);

The unit-converted vector should now appear.

The issue here is somewhat like the rtablesize issue that I wrote about an hour ago in the sense that the computations are already done but you're not seeing the results; however, in this case the issue doesn't only affect the display. (It took me many years to fully understand the rtable_eval command, so you'll likely need to just accept that it's the "magic command" that does what you want in this case.)

What you described is indeed an efficient (although somewhat mysterious) way to fill a matrix. It doesn't matter whether the entries are floats or any other type. The key step is to use the very mysterious command rtable_eval on the row vector; it changes the symbolic variables (A, B, C, etc.) in the row vector to their assigned values. Like this:

restart:
OutVec:= <A, B, C>^%T: #row vector
OutMat:= Matrix((9, numelems(OutVec)), datatype= anything):
#The datatype could be hfloat or anything else or omitted entirely.

MyFuncs:= proc(k)
    :-A:= (k-1)^2;
    :-B:= k^2;
    :-C:= (k+1)^2;
    rtable_eval(OutVec)
end proc
:
for k to upperbound(OutMat)[1] do
    OutMat[k]:= MyFuncs(k)
od
:
OutMat;

Whether you send this matrix to Excel afterwards is irrelevant; that'd be handled exactly like sending any matrix to Excel, e.g., as shown by @dharr .

I can't tell you how to do it with context-menu-based commands such as you appear to be using, but these text-based commands do it:

convert~(<20., 30., 40.>*Unit(MPa), units, psi);
convert~(<150., 300., 600.>*Unit(psi), units, MPa);

Note the ~ after convert, which causes the command to be applied elementwise to the vector. It works exactly the same for a matrix, array, list, set, or table.

What you describe should be fairly easy because Maple (since Maple 2018) has an on-board Python interpreter and a small package of commands to support it. See help page ?Python.

It's easier to write a procedure for this that works for an arbitrary number of arguments than it is to write one specific to 4 arguments. In the procedure below, you'll need to fill in the ...'s.

biggest4:= proc(S::seq(realcons)) #i.e., the arguments will be a SEQuence of REAL CONStants
local 
    M:= S[1], #Initialize max to 1st argument (arbitrarily).
    x #any of the arguments
;
    #Compare every argument to the max computed so far and adjust it if needed:
    for x in S do
        #The "is" command is sometimes needed to compare complicated symbolic
        #expressions that represent real numbers:
        if is(x > ...) then ...:= ... fi 
    od;
    M #procedure's return value
end proc
:
biggest4(-1/3, 0, 7/3, -2);

If you omit the is, this'll still work for these example numbers.