Creating a loop with for loop

How do you loop A11-A33 correctly?

function [determinant , inverse ] = invanddet3by3(A)
A11 =  invanddet2by2sol(A([2,3], [2,3])); % Cofactors 3x3 matrix A
A12 = -invanddet2by2sol(A([2,3], [1,3]));
A13 =  invanddet2by2sol(A([2,3], [1,2]));
A21 = -invanddet2by2sol(A([1,3], [2,3]));
A22 =  invanddet2by2sol(A([1,3], [1,3]));
A23 = -invanddet2by2sol(A([1,3], [1,2]));
A31 =  invanddet2by2sol(A([1,2], [2,3]));
A32 = -invanddet2by2sol(A([1,2], [1,3]));
A33 =  invanddet2by2sol(A([1,2], [1,2]));
D   = [A11 A12 A13; A21 A22 A23; A31 A32 A33]; % Adju Matrix
determinant = A(1,1) * A11 + A(1,2) * A12 + A(1,3) * A13; % Deter of A
if determinant == 0
    inverse=[];
else
    inverse = D' / determinant;  % Inv of A
end
end

[Rik]

What do you mean?

[Miguel Anliker]

I have to create a loop so that code lines 2-10 are simplified

[Stephen23]

"I have to create a loop so that code lines 2-10 are simplified"

Then don't use numbered variables.

Using numbered variables is a sign that you are doing something wrong.

https://www.mathworks.com/matlabcentral/answers/489270-return-many-value-in-function#answer_399959

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[JESUS DAVID ARIZA ROYETH]

solution:

function [determinant , inverse ] = invanddet3by3(A)
D=zeros(3);
s=[2 3; 1 3; 1 2];
for k=1:size(D,1)
    for j=1:size(D,2)
        D(k,j)=(-2*mod(k+j,2)+1)*invanddet2by2sol(A(s(k,:),s(j,:)));
    end
end
determinant=sum(A(1,:).*D(1,:));
if determinant == 0
    inverse=[];
else
    inverse = D' / determinant;  % Inv of A
end
end

[Miguel Anliker]

Just a small question :

Can you explain a bit more in detail how you got the loop function "for" and if possible simplify it ?

Thank you Jesus!

[JESUS DAVID ARIZA ROYETH]

yes, it is even possible to eliminate the for cycle, and receive any dimension input, basically if the sum of the row and the column gives odd then it is multiplied by -1 otherwise multiplied by 1 ((-2 * mod (k + j, 2) +1 ) -> 2x-1 --> if x==1 then y=1 else if x==0 then y=-1), I also noticed that the number 1 corresponds [2,3], the number 2 corresponds [1,3] and so on

[Rik]

Comment posted as answer by Miguel Anliker:

Thank you for the explanation.