Caculating the sum of elements and building a vector and a matrix with letters

I created the following matrix:

Here's the code:

n = input('enter n\n');

A=[sym('m%d', [n 1]) sym('x%d', [n 1]) sym('y%d', [n 1]) sym('z%d', [n 1])];

I need to build the following vector and matrix:

Once with using loops and once without using loops. I don't succeed to create a sum of elements, that are letters. How Can I build those arrays?

Answer 1

[John BG]

Hi Lee

this is John BG <mailto:[e-mail] [e-mail]>

1.-

Simulating data

n=12
m=randi([1 10],n,1);
x=randi([-13 13],n,1);
y=randi([-13 13],n,1);
z=randi([-13 13],n,1);

2.-

Your matrix

A=[m x y z]

3.-

the v vector

v=[sum(m.*x) sum(m.*y) sum(m.*z)]

it's the same as

v=[sum(A(:,1).*A(:,2)) sum(A(:,1).*A(:,3)) sum(A(:,1).*A(:,4))]

4.-

Regarding the Inertia matrix I, once you have keyed in the matrix A with the couple lines shown in your question

n = input('enter n\n');
A=[sym('m%d', [n 1]) sym('x%d', [n 1]) sym('y%d', [n 1]) sym('z%d', [n 1])];

the generatrion of I is more compact using vectors m x y and z

To get these vectors one can easily extract m x y z with

m=A(:,1);
x=A(:,2);
y=A(:,3);
z=A(:,3);

then, note that it may be the case that you need I to be 3D so you can address the Inertia moment for any given i-th layer

I0=zeros(3,3,n)

% 1st row of I
I0(1,1,:)=m.*(x).^2
I0(2,1,:)=m.*x.*y
I0(3,1,:)=m.*x.*z

% 2nd row of I
I0(1,2,:)=m.*x.*y
I0(2,2,:)=m.*(y).^2
I0(3,2,:)=m.*z.*y

% 3rd row of I
I0(1,3,:)=m.*x.*z
I0(2,3,:)=m.*y.*z
I0(3,3,:)=m.*(z).^2

5.-

then one can obtain the sum matrix I just adding along 3rd dimension of I0

I=sum(I0,3)

f you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>

[Lee Cohen]

Thank you so much, but I also need another solution using loops.

[John BG]

indeed, no problem. Before I start with the loops, it shouldn't take long, would it be possible for you to mark as accepted the compact answer?

Actually the loops take more lines that the solution above supplied, do you really need them?

Regards

John BG

<mailto:[e-mail] [e-mail]>

[Lee Cohen]

Yes, this is the part of the answer that I used: v=[sum(m.*x) sum(m.*y) sum(m.*z)]

I know it takes more lines with using loops, but it is a part of my assignment.

[John BG]

no problem, here we go:

1.-

 generating data

n=12
m=randi([1 10],n,1);
x=randi([-13 13],n,1);
y=randi([-13 13],n,1);
z=randi([-13 13],n,1);
A=[m x y z]   % A, the matrix you start the question with
m=A(:,1);
x=A(:,2);
y=A(:,3);
z=A(:,3);

2.-

the equivalent to

% v=[sum(m.*x) sum(m.*y) sum(m.*z)]

.

is

.

mx=0;
for k=1:1:numel(m)
    mx=mx+m(k)*x(k);
end

my=0;
for k=1:1:numel(m)
    my=my+m(k)*y(k);
end

mz=0;
for k=1:1:numel(m)
    mz=mz+m(k)*z(k);
end

v_sum=[mx my mz]

.

or

.

p=[x y z];
v_sum=[0 0 0]
for k=1:1:numel(m)
    v_sum=v_sum+m(k)*p(k,:);
end

3.-

the equivalent to

v=[sum(A(:,1).*A(:,2)) sum(A(:,1).*A(:,3)) sum(A(:,1).*A(:,4))]

is

I_sum=zeros(3);
for k=1:1:numel(m)
    I_sum=I_sum+m(k)*[(x(k))^2 x(k)*y(k) x(k)*z(k);x(k)*y(k) (y(k))^2 y(k)*z(k); x(k)*y(k) z(k)*y(k) (z(k))^2];
end

.

or

.

I_sum=zeros(3);
for k=1:1:numel(m)
    D=zeros(3);
    p1=p(k,:);
    for s1=1:1:3
        for s2=1:1:3
            D(s1,s2)=p1(s1)*p1(s2);
        end
    end
    I_sum=I_sum+m(k)*D;
end

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>

[Lee Cohen]

Thank you so much!

[John BG]

Any time, happy to help :)