# 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.

Thahk you for the helpful answer. But I have to avoid loops here.
Steven Lord on 9 Oct 2019
Why? Because that's a condition of your assignment or because you have heard that loops are slow in MATLAB? If the latter, that's become less and less accurate (when the loop is written well) as time has progressed and MATLAB has improved.