why in this function they use empty square brackets?

and how exactly this is solving the determinant, I don't completley understand what they did here/
sorry if it's a dumb question I'm new to coding and to matlab.
function det = myDet(A)
if isscalar(A)
det = A;
return
end
det = 0;
top_row = A(1,:);
A(1,:) = [];
for i = 1:size(A,2) % number of columns
A_i = A;
A_i(:,i) = [];
det = det+(-1)^(i+1)*top_row(i)*myDet(A_i);
end
end

 Risposta accettata

It is not a dumb question at all, because the syntax used by MATLAB to delete a row or column of an array may not be obvious. But the best way, as others have said, is to try it out. See what happens to the array when you do that. Getting your hands dirty is the best way to learn.
A = magic(3)
A = 3×3
8 1 6 3 5 7 4 9 2
% first, I'll delete the second row of A
A(2,:) = []
A = 2×3
8 1 6 4 9 2
% Next, delete the third column of that matrix.
A(:,3) = []
A = 2×2
8 1 4 9

Più risposte (2)

In this context, it appears to be deleting elements from A.
Try it out in MATLAB to see what it is doing.
A = rand(3)
A = 3×3
0.0943 0.6034 0.8591 0.6091 0.1028 0.2550 0.9727 0.2310 0.9515
A(1,:)=[]
A = 2×3
0.6091 0.1028 0.2550 0.9727 0.2310 0.9515
Jan
Jan il 13 Nov 2021
Modificato: Jan il 13 Nov 2021
A(i,:) = [];
This deletes the i.th row of the matrix A.
Example:
A = [1,2,3; 4,5,6; 7,8,9];
A(1, :) = []
% A = [4,5,6; 7,8,9]

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