TheKingDanQc

Unfortunately I did not understand what you wrote. Here are the other 2 ways.

The easiest way to write such a procedure - it's just  to use a ready formula and write it as a procedure-function:

det := A::Matrix -> A[1, 1]*A[2, 2]*A[3, 3] - A[1, 1]*A[2, 3]*A[3, 2] - A[1, 2]*A[2, 1]*A[3, 3] + A[1, 2]*A[2, 3]*A[3, 1] + A[1, 3]*A[2, 1]*A[3, 2] - A[1, 3]*A[2, 2]*A[3, 1] :

Example of use:

A := <a1, b1, c1; a2, b2, c2; a3, b3, c3>;

det(A); 

The second method uses the definition of the determinant of any 3x3 matrix as the sum of the summands of the form  

sign((j-i)*(k-i)*(k-j))*A[i, 1]*A[j, 2]*A[k, 3]

 taken over all permutations of  {i, j, k} :

det1 := proc (A::Matrix)

local S, Rows, i, j, k;

S := 0; Rows := {1, 2, 3};

for i to 3 do

for j in Rows minus {i} do

for k in Rows minus {i, j} do

S := S+sign((j-i)*(k-i)*(k-j))*A[i, 1]*A[j, 2]*A[k, 3]

od; od; od;

S;

end proc:

The procedure  det1  is easily generalized to the case of a determinant of order 4, if you add 1 more loop.

The solution easy to get by  fsolve  command if we make a little preparatory work. The system can be reduced to two equations with two unknowns, if we subtract the second equation from the first equation. By  plots[implicitplot]  we find the initial approximations for  T[i]  and  T[r] , and the initial value for  H  we find by solving the first equation:

restart:

T[l]:= 20:

C[p]:= 3779:

rho:= 1026:

nu:= 0.004/60:

P:= 1000:

lign1:= H = 40.09 + 14.774*ln(T[i]-T[l])+(2.28+0.338*ln(T[i]-T[l]))*T[l]:

lign2:= H = -nu*rho*C[p]*ln(1-(T[i]-T[r])/(T[i]-T[l])):

lign3:= P = nu*rho*C[p]*(T[i]-T[r]):

plots[implicitplot]([lign1-lign2, lign3], T[i]=20..50, T[r]=20..50, color=red, thickness=2, gridrefine=3);

fsolve({lign||(1..3)}, {T[i]=30, T[r]=25, H=solve(eval(lign1,T[i]=30))});

An example of using numerical solution:

restart;

dsys := {diff(x(t),t)=y(t),diff(y(t),t)=-x(t),x(0)=1,y(0)=0}:

Sol:=dsolve(dsys,numeric, output=listprocedure);

x:=rhs(Sol[2]);

y:=rhs(Sol[3]);

fsolve(sin(x(t))+x(t)*y(t)^3-0.3=0, t=0..2);  # The solution of a nonlinear equation in a specific range

plot(sin(x(t))+x(t)*y(t)^3-0.3, t=0..2, color=red, thickness=2);  # Visualization

Edited. 

You can exactly to solve your equation  Eq  and then to define the function  U(x,t):

restart;

Eq:=diff(u1(z),z)=9/4*(84*(-z)^(11/2)-8*z^7+540*(-z)^(5/2)+324*z^4-324*z)*u1(z)/((z^2+3*sqrt(-z))*(2*(-z)^(3/2)+3)*(8*z^6+66*(-z)^(9/2)-189*z^3+216*(-z)^(3/2)+81)):

dsolve({Eq, u1(-2)=1});  # Exact solution

assign(%);

u0(z):=z^2/3+sqrt(-z):

U:=unapply(eval(u0(z)+t*u1(z), z=x/t),x,t);

plot(U(x,1), x=-2..0, color=red, thickness=2);  # The plot  U(x,t)  for  t=1

1) You forgot to call  CurveFitting  package. Instead  PolynomialInterpolation  should be  CurveFitting[PolynomialInterpolation]

2) f=-z  is the equation of degree 5 with 2 parameters. Even if there were no parameters, such equations in the general case can not be solved symbolically in terms of its coefficients (in radicals). See  wiki  and  wiki . In the best case, you can solve it numerically (if you specify parameter values).

da := [[1,m],[2,m2],[3,1],[4,2],[5,3],[6,4]];

f := CurveFitting[PolynomialInterpolation](da, z);

solution := solve(f=-z, z);

fsolve(eval(f=-z, [m=1,m2=2]), complex); # Numerical solution for specific values of parameters

Suppose you are looking for the lowest four-digit prime number whose sum of digits is divisible by 17. Write the program as a procedure without parameters:

restart;

Check:=proc()

local i, j, k, l, b;

for i from 1 to 9 do

for j from 0 to 9 do

for k from 0 to 9 do

for l from 0 to 9 do

b:=1000*i+100*j+10*k+l;

if isprime(b) and irem(i+j+k+l, 17)=0 then return i, j, k, l  fi;

od: od: od: od:

end proc:

Check();

                                       1, 0, 9, 7

Addition.  Using a procedure also has the advantage that all of the variables, to which you give any values in the procedure's body by default  will be local, and outside the body of the procedure they are ordinary symbols:

P:=proc()

local i, j;

i:=1; j:=2;

end proc:

i;  j;

                                 i

                                 j

See help on  Student[LinearAlgebra][GaussianEliminationTutor]  command.

If I understand you're looking for minimax optimization. DirectSeach  package provides a slightly better solution:

DirectSearch[GlobalOptima](max(abs(pa^2 + 2*pa*po - 0.44), abs(pb^2 + 2*pb*po - 0.10), abs(po^2 - 0.42), abs(2*pa*pb - 0.04)), [pa = 0.279246445735439, pb = 0.0728480349005200, po = 0.648114970425358]);

0.000685054125967331, [pa =0 .27924642145279, pb = 0. 07284794181838, po =0 .64811497042536], 925]

because  0.000685054125967331<0.000685109649562879

Sys := {diff(x(t),t) = y(t)-2*(x(t))^2, diff(y(t),t) = -8*(x(t))^3+4*x(t)*y(t)-5*(x(t))^4-(x(t))^5};

DETools[phaseportrait](Sys, [x(t), y(t)], t=-1..1, [[x(0)=1, y(0)=0]]);

R:=solve(f, x, explicit):

`*`(seq(x-R[i], i=1..4));

I pretty often use it. The operator  ~  seems to appear beginning with Maple 15 or Maple 16. It replaces not only  map  operator, but also  zip  operator. Here is an example in which you want to equate each element of the first list to the square of the corresponding element of second list:

X:=[x,y,z]:  A:=[a,b,c]:

zip(`=`, X, map(t->t^2,A))  # First way

X=~A^~2;  # Second way

                                 [x = a^2, y = b^2, z = c^2]

                                 [x = a^2, y = b^2, z = c^2]

This operator provides a more compact (short) syntax. I do not know his other advantages.

This works in one session until  restart  command is not used.

local I;

I:=2;

I+1;

                                      I := 2

                                         3

1) See  LinearAlgebra[HilbertMatrix]  command.

2) Your matrix  (2*2 matrix which its transpose is egual to its inverse)  is a rotation matrix or composition of a rotation and a reflection:  

Matrix([[cos(alpha), -sin(alpha)], [sin(alpha), cos(alpha)]]);  # A rotation matrix

Matrix([[1,0], [0,-1]]) . Matrix([[cos(alpha), -sin(alpha)], [sin(alpha), cos(alpha)]]);  # Composition of a rotation and a reflection

We have a system with 27 unknowns and 54 variables, that is 27 variables are parameters. If you give some values to these parameters, fsolve  easy solves the system. Replace the last line of your code by

ABC := fsolve(eval(map(t->Equate(op(t))[], [eq2 = AA1, eq3 = AA2, eq4 = B2]), {aaa1 = 1, aaa2 = 2, aaa3 = 3, aaa4 = 4, aaa5 = 5, aaa6 = 6, aaa7 = 7, aaa8 = 8, aaa9 = 9, aab1 = 10, aab2 = 11, aab3 = 12, aab4 = 13, aab5 = 14, aab6 = 15, aab7 = 16, aab8 = 17, aab9 = 18, aabb1 = 19, aabb2 = 20, aabb3 = 21, aabb4 = 22, aabb5 = 23, aabb6 = 24, aabb7 = 25, aabb8 = 26, aabb9 = 27}));

{a1 = 1.461172682, a2 = 0.4611726819, a3 =0 .4611726819, a4 = 8.077654636, a5 = 9.077654636, a6 = 8.077654636, a7 = -8.538827318, a8 = -8.538827318, a9 = -7.538827318, b1 = -3.221842064, b2 = -2.273083473, b3 = -2.324324882, b4 = 10.46946092, b5 = 8.571943737, b6 = 8.674426555, b7 = -7.673550001, b8 = -6.724791410, b9 = -6.776032819, c1 = 31.22184206, c2 = 31.27308347, c3 = 32.32432488, c4 = 20.53053908, c5 = 23.42805626, c6 = 24.32557344, c7 = 41.67355000, c8 = 41.72479141, c9 = 42.77603282}

This is an answer to the second question. The answer to the first question is obvious.

Basic variables corresponds to the maximum set of linearly independent columns of the leading matrix of the system. The remaining variables are the free variables, which we consider as parameters. Selection of basic variables is ambiguous. I think the default Maple selects the first set in lexicographical order. For example in solving of  {x-y + z = 2, 2 * x + y-z = 3}  Maple expresses  x  and  y  through  z . But for the system  {x-y+z=2, 2*x-2*y-3*z=5}  x  and  z  are expressed through  y , because the first 2 columns are linearly dependent. 

You can choose yourself the basic variables (if the corresponding columns are linearly independent) as in the example:

solve({x-y+z=2, 2*x+y-3*z=5}, {x,z});

                                   {x = 11/5+2/5*y, z = -1/5+3/5*y}