# How to calculate determinant of matrices without loop?

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I am new to Matlab and this might seem very easy.
I have 2 matrices:
a = [1 1 1; 2 2 2 ; 3 3 3 ; 4 4 4 ; 5 5 5];
b = [4 4 4; 3 2 4 ; 1 5 7 ; 4 3 8 ; 2 4 7];
I wanted to calculate the determinant of each row of the two matrices added by a row of ones (a 3*3 matrix), and put all the determinants in another array. For example, first determinant (d(1)) would be from this matrix:
1 1 1
4 4 4
1 1 1
and the second one (d(2)) would be from this matrix:
2 2 2
3 2 4
1 1 1
and so on...
When I try this:
m = size(a,1);
ons = ones(m,3);
d = det([a(:,:) ; b(:,:) ; ons(:,:)]);
I get this error:
Error using det
Matrix must be square.
How can I calculate all the determinants at once without using loop?

James Tursa on 9 Oct 2019
Well, for this particular example, the result is simply
d = [0,0,0,0,0];
because all the rows of a are multiples of [1 1 1]
You are right. But when I use:
d = arrayfun(@(x)det([a(x,:);b(x,:);ones(1,3)]),1:length(a));
the result is:
1.0e-15 *
0 0 -0.3331 0 0.8327
Rik on 10 Oct 2019
There are 16 orders of magnitude between input and output. That is fairly close to eps, so this could be a float rounding error (and it is).

Bruno Luong on 10 Oct 2019
a(:,1).*b(:,2)-a(:,2).*b(:,1)-a(:,1).*b(:,3)+a(:,3).*b(:,1)+a(:,2).*b(:,3)-a(:,3).*b(:,2)

Bruno Luong on 10 Oct 2019
Some timing
a=rand(1e6,3);
b=rand(1e6,3);
tic
d=arrayfun(@(x)det([a(x,:);b(x,:);ones(1,3)]),(1:length(a))');
toc % 5.066323 seconds.
tic
A = reshape(a.',1,3,[]);
B = reshape(b.',1,3,[]);
ABC = [A; B];
ABC(3,:,:) = 1;
d = zeros(size(a,1),1);
for k=1:size(a,1)
d(k) = det(ABC(:,:,k));
end
toc % Elapsed time is 1.533522 seconds.
tic
d = a(:,1).*b(:,2)-a(:,2).*b(:,1)-a(:,1).*b(:,3)+a(:,3).*b(:,1)+a(:,2).*b(:,3)-a(:,3).*b(:,2);
toc % Elapsed time is 0.060121 seconds.
I keep writing since day one that ARRAYFUN is mostly useless when speed is important.

David Hill on 9 Oct 2019
d=arrayfun(@(x)det([a(x,:);b(x,:);ones(1,3)]),1:length(a));

Rik on 9 Oct 2019
This is probably the solution OP is looking for, but I want to add that this still contains a loop. It just moves the loop internally, while adding a layer of complexity with the anonymous function.
I know it is an internal loop, but this is what I was looking for and it is working well. Thank you.
Bruno Luong on 10 Oct 2019
You have accepted the worse answer in term of runing time, see the tic/toc I made below

Steven Lord on 9 Oct 2019
This line of code:
d = det([a(:,:) ; b(:,:) ; ons(:,:)]);
Stacks all of a on top of all of b, and stacks that on top of all of ons. It then tries to take the determinant of that array. Let's see the matrix you created.
d = [a(:,:) ; b(:,:) ; ons(:,:)]
d =
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
4 4 4
3 2 4
1 5 7
4 3 8
2 4 7
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
The easiest way to accomplish what you want is a for loop that iterates through the rows of the a and b matrices. It's hard to give an example of the technique on your data that doesn't just give you the solution (which I'd prefer not to do, since this sounds like a homework assignment.) So I'll just point to the array indexing documentation. You want to access all of one of the rows of a (and all of the same row of b) and use that accessed data to create the d matrix for that iteration of the for loop. Each iteration will have a different d.