Creating a loop with for loop
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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 个评论
Rik
2019-11-7
What do you mean?
Miguel Anliker
2019-11-7
Stephen23
2019-11-7
"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.
Rena Berman
2019-12-12
(Answers Dev) Restored edit
采纳的回答
JESUS DAVID ARIZA ROYETH
2019-11-7
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 个评论
Miguel Anliker
2019-11-7
JESUS DAVID ARIZA ROYETH
2019-11-7
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
2019-11-7
Thank you for the explanation.
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