I need help trying to code an mn by mn A matrix to be use to calculate temperature

I have a hard coded version of what the code sure do but it is returning the warning that the matrix is singular. I have started coding for an m by n however I getting confuse on how to code using if statements. The problem have the conditions that the left and bottom side of the red figure is insulated and the right and top side have the different h values and a cooling temperature of 25 deg C. I have provided an figure to help visual the problem:

%% Input 
T_inf = 25; % C
kx = 5; % W/mk 
ky = 10; % W/mk
q_dot = 200000; % W/m^3 
hy = 10; % W/m^2K 
hx = 100; %W/m^2k
% nodes
m = 5;
n = 5;
%Grid Setup 
X = .025; %m 
Y = .1; %m
% Spacing
dx = X/m; 
dy = Y/n;
% Create A matrix 
s = m*n;
A = zeros(s,s);
% Create b matrix 
b = zeros(s,1);
%% Mohen 
A11 = -3.*((hx.*dx./2) + (kx.*dy./(2.*dx)) + (ky.*dx./(2.*dy)));
A12 = kx.*dy./2;
A13 = ky.*dx./2;
A21 = -4 .* (kx.*dx + kx.*dy./(2.*dx) + ky.*dx./dy + kx.*dy./(2.*dx));
A22 = kx.*dy./(2.*dx);
A23 = ky.*dx./dy;
A24 = kx.*dy./(.2*dx);
A31 = -4.*(hx.*dx./2 + hy.*dy ./ 2 + ky.*dx./(2.*dx)+kx.*dy./(2.*dx));
A32 = ky.*dx./(2.*dy);
A33 = kx.*dy./(2.*dx);
A41 = -3.*(ky.*dx./(2.*dy)+ kx .* dy./dx + ky.*dx./(2.*dy));
A42 = ky.*dx./(2.*dy);
A43 = kx.*dy./dx;
A44 = ky.*dx./(2.*dy);
A51 = -4.*(ky.*dx./dy + kx.*dy./dx + ky.*dx./dy + ky.*dy./dx);
A52 = ky.*dx./dy;
A53 = kx.*dy./dx;
A54 = ky.*dx./dy;
A55 = kx.*dy./dx;
%% Alice
A61 = -4.*(ky.*dx./(2.*dy)+hy.*dy+ky.*dx./(2.*dy)+kx.*dy./dx); 
A62 = ky*dx/2/dy;
A63 = ky*dx/2/dy; 
A64 = kx*dy/dx; 
A71 = -2*(ky*dx/2/dy+kx*dy/2/dx);
A72 = ky*dx/2/dy;
A73 = kx*dy/2/dx; 
A81 = -3*(ky*dx/dy+kx/2/dx);
A82 = ky*dx/dy; 
A83 = kx*dy/2/dx;
A84 = kx*dy/2/dx;
A91 = -3*(ky*dx/2/dy+hy*dy/2+kx*dy/2/dx);
A92 = ky*dx/2/dy;
A93 = kx*dy/2/dx;
%% Travion 
b1 = -hx*(dx/2)*T_inf - q_dot*(dx/2)*(dy/2);
b2 = -hx*dx*T_inf - q_dot*dx*(dy/2);
b3 = -hx*(dx/2)*T_inf - q_dot*(dx/2)*(dy/2) - hy*(dy/2)*T_inf;
b4 = -q_dot*(dx/2)*dy;
b5 = -q_dot*dx*dy;
b6 = -q_dot*(dx/2)*dy-hy*dy*T_inf;
b7 = -q_dot*(dx/2)*(dy/2);
b8 = -q_dot*dx*(dy/2);
b9 = -q_dot*(dx/2)*(dy/2)-hy*(dy/2)*T_inf;
%% Hard Code
A(1,1) = A11; 
A(1,2) = A12; 
A(1,6) = A13;
A(2,2) = A21; 
A(2,3) = A22; 
A(2,7) = A23; 
A(2,1) = A24;
A(3,3) = A21; 
A(3,4) = A22; 
A(3,8) = A23; 
A(3,2) = A24;
A(4,4) = A21; 
A(4,5) = A22; 
A(4,9) = A23; 
A(4,3) = A24;
A(5,5) = A31; 
A(5,10) = A32; 
A(5,4) = A33;
A(6,6) = A41; 
A(6,1) = A42; 
A(6,7) = A43; 
A(6,11) = A44;
A(11,11) = A41; 
A(11,6) = A42; 
A(11,12) = A43; 
A(11,16) = A44;
A(16,16) = A41; 
A(16,11) = A42; 
A(16,17) = A43; 
A(16,21) = A44;
A(7,7) = A51; 
A(7,2) = A52; 
A(7,8) = A53; 
A(7,12) = A54; 
A(7,6) = A55;
A(8,8) = A51;
A(8,3) = A52;
A(8,9) = A53;
A(8,13) = A54;
A(8,7) = A55;
A(9,10) = A51;
A(9,5) = A52;
A(9,10) = A53;
A(9,14) = A54;
A(9,8) = A55;
A(12,12) = A51;
A(12,7) = A52;
A(12,13) = A53;
A(12,17) = A54;
A(12,11) = A55;
A(13,13) = A51;
A(13,8) = A52;
A(13,14) = A53;
A(13,18) = A54;
A(13,12) = A55;
A(14,14) = A51;
A(14,9) = A52;
A(14,15) = A53;
A(14,19) = A54;
A(14,13) = A55;
A(17,17) = A51;
A(17,12) = A52;
A(17,18) = A53;
A(17,22) = A54;
A(17,16) = A55;
A(18,18) = A51;
A(18,13) = A52;
A(18,19) = A53;
A(18,23) = A54;
A(18,17) = A55;
A(19,19) = A51;
A(19,14) = A52;
A(19,20) = A53;
A(19,24) = A54;
A(19,18) = A55;
A(10,10) = A61;
A(10,5) = A62; 
A(10,15) = A63; 
A(10,9) = A64; 
A(15,15) = A61;
A(15,10) = A62; 
A(15,20) = A63; 
A(15,14) = A64;
A(20,20) = A61;
A(20,15) = A62; 
A(20,25) = A63; 
A(20,19) = A64;
A(21,21) = A71; 
A(21,16) = A72; 
A(21,22) = A73; 
A(22,22) = A81; 
A(22,17) = A82; 
A(22,23) = A83;
A(22,21) = A84;
A(23,23) = A81; 
A(23,18) = A82; 
A(23,24) = A83;
A(23,22) = A84;
A(22,24) = A81; 
A(22,19) = A82; 
A(22,25) = A83;
A(22,23) = A84;
A(25,25) = A91; 
A(25,20) = A92; 
A(25,24) = A93;
b = [b1; b2; b2; b2; b3; b4; b5; b5; b5; 
    b6; b4; b5; b5; b5; b6; b4; 
    b5; b5; b5; b6; b7; b8; b8; b8; b9;];
T = A\b;
%% Actual Code for an N x M 
[C,D] = sizes(A);
for i = 1:c
    for j = 1:d
        if i = 2:c-1 

Incorrect use of '=' operator. Assign a value to a variable using '=' and compare values for equality using '=='.

            if j = 2:d-1 
                A(i,j) = A51;
                A(i,j-d) = A52;
                A(i,j+1) = A53;
                A(i,j+d) = A54;
                A(i,j-1) = A55;
            end
            
            
        end
    end

Answer 1

[Walter Roberson]

if i >= 2 && i <= c-1
    if j >= 2 && j <= d-1