How to calculate determinant of matrices without loop?

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?

[Rik]

Why do you want to avoid a loop? It seems the easiest solution for your needs.

[John D'Errico]

What you tried to do will obviously cause an error, for good reason. However...

The determinant of a single row of a matrix has no meaning in mathematics. So you need to explain more clearly what you want. Or perhaps more accurately, what you think you want.

[Walter Roberson]

Are you trying to use the Minor Expansion Formula for determinant?

[Hadi Ghahremannezhad]

I added explaination in the question.

[James Tursa]

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]

[Hadi Ghahremannezhad]

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]

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).

Answer 1

[Bruno Luong]

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]

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.

[Hadi Ghahremannezhad]

Great answer.

Answer 2

[David Hill]

d=arrayfun(@(x)det([a(x,:);b(x,:);ones(1,3)]),1:length(a));

[Rik]

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.

[Hadi Ghahremannezhad]

I know it is an internal loop, but this is what I was looking for and it is working well. Thank you.

[Bruno Luong]

You have accepted the worse answer in term of runing time, see the tic/toc I made below

Answer 3

[Steven Lord]

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.

[Hadi Ghahremannezhad]

Thahk you for the helpful answer. But I have to avoid loops here.

[Steven Lord]

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.

[Hadi Ghahremannezhad]

Both. This is a small part of an assignment. It says always use vectorization instead of loops because of higher speed. The assignment is about mesh processing and I am trying to calculate the area of each face (traingle) where I have the 3 points of each triangle.