Как найти произведение каждого столбца двумерного массива

Упрощаем, пишем функцию для поиска произведения членов списка, конкретно для данной задачи для одного столбца

import math

def column_mul(matrix, number):
    """Возвращает произведение членов столбца с номером number"""
    column = [row[number] for row in matrix] # столбец с номером number
    return math.prod(column)

Далее используем данную функцию для количества столбцов и храним результат в переменной mul_all_columns, а затем используем метод min

mul_all_columns = []
for i in range(len(matrix)):
    mul_all_columns.append(column_mul(matrix, i))

minimum = min(mul_all_columns)

либо можно сократить до одной строки

minimum = min(column_mul(matrix, i) for i in range(len(matrix)))

также можно и максимум найти

maximum = max(column_mul(matrix, i) for i in range(len(matrix)))

uses crt;
const nmax=20;
type matr=array [1..nmax,1..nmax] of integer;
     mas=array [1..nmax] of integer;
var a: matr;
    a1,a2,a3,a4: mas;
    n,m: byte;
    
procedure Vvod (var z: matr; var ri,rj: byte);
var i,j: byte;
begin
repeat
write('Kol-vo strok, ne menee 4, ne bolee ',nmax,', n='); readln(ri);
until ri in [4..nmax];
repeat
write('Kol-vo stolbzov, ne bolee ',nmax,', n='); readln(rj);
until rj in [1..nmax];
writeln('Vvedite elementy matrizy:');
for i:=1 to ri do
    for j:=1 to rj do
        begin
        write('a[',i,',',j,']='); readln(z[i,j]);
        end;
end;
 
procedure Vivod (z: matr; ri,rj: byte);
var i,j: byte;
begin
writeln('Ischodnaja matriza');
for i:=1 to ri do
    begin
    for j:=1 to rj do
        write(z[i,j]:4);
    writeln;
    end;
end;
 
procedure Vivod1M (c: char; z: mas; rj: byte);
var j: byte;
begin
writeln('Massiv ',c);
for j:=1 to rj do
    write(z[j]:4);
writeln;
end;
 
procedure MakeArray (ns: byte; x: matr; var y: mas; rj: byte);
var j: byte;
begin
for j:=1 to rj do
    y[j]:=x[ns,j];
end;
 
function Jmin (z: mas; rj: byte): byte;
var j,jm: byte;
begin
jm:=1;
for j:=1 to rj do
    if z[j]<z[jm] then jm:=j;
Jmin:=jm;
end;
 
procedure Zamena (var z: mas; rj: byte);
var j: byte; g: integer;
begin
g:=z[Jmin(z,rj)];
if Jmin(z,rj)<rj then
   for j:=Jmin(z,rj)+1 to rj do
       z[j]:=g;
end;
 
begin
Vvod (a,n,m);
clrscr;
Vivod (a,n,m);
MakeArray (1,a,a1,m);
MakeArray (2,a,a2,m);
MakeArray (3,a,a3,m);
MakeArray (4,a,a4,m);
writeln('Massivy do preobrazovania');
Vivod1M ('1',a1,m);
Vivod1M ('2',a2,m);
Vivod1M ('3',a3,m);
Vivod1M ('4',a4,m);
Zamena (a1,m);
Zamena (a2,m);
Zamena (a3,m);
Zamena (a4,m);
writeln('Massivy posle zameny');
Vivod1M ('1',a1,m);
Vivod1M ('2',a2,m);
Vivod1M ('3',a3,m);
Vivod1M ('4',a4,m);
readln;
end.

Start with the declaration: make sure that your program works with m×n matrix, not simply a 3×3 matrix. Since m and n have limits of 10 and 20, and because you must add an extra row and a column to the result, the declaration should be

int a[11][21];

You also need to declare m and n, have end-user enter them, and validate them to be within their acceptable ranges:

int m, n;
cin >> m >> n;
... // Do the validation

Now you can rewrite your loops in terms of m and n, rather than using 3 throughout your code.

With these declarations in place, you can total the numbers in place, i.e. for each row r you would write

for (int i = 0 ; i != n ; i++) {
    a[r][n+1] += a[r][i];
}

Similarly, you would compute the product (don’t forget to start it with the initial value of 1, not 0).

At the end you would print an (m+1)×(n+1) matrix to complete the task.