Creating a loop with for loop
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Miguel Anliker
on 7 Nov 2019
Commented: Rena Berman
on 12 Dec 2019
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
4 Comments
Stephen23
on 7 Nov 2019
"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.
Accepted Answer
JESUS DAVID ARIZA ROYETH
on 7 Nov 2019
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
3 Comments
JESUS DAVID ARIZA ROYETH
on 7 Nov 2019
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
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