Generate Matrix with MATLAB - 65 Characters

[DryRun]

I have attached the problem to this post. My attempt at the first part (i used 'm' instead of 'l' as it's less confusing, since the latter resembles the digit '1') and here is my script:

A = zeros(10);
for k=1:10 
  for  m=1:10
    A(k,m)  =  sin(k)*cos(m); 
  end 
end 
 
A

The answer:

A =

0.4546 -0.3502 -0.8330 -0.5500 0.2387 0.8080 0.6344 -0.1224 -0.7667 -0.7061
0.4913 -0.3784 -0.9002 -0.5944 0.2579 0.8731 0.6855 -0.1323 -0.8285 -0.7630
0.0762 -0.0587 -0.1397 -0.0922 0.0400 0.1355 0.1064 -0.0205 -0.1286 -0.1184
-0.4089 0.3149 0.7492 0.4947 -0.2147 -0.7267 -0.5706 0.1101 0.6895 0.6350
-0.5181 0.3991 0.9493 0.6268 -0.2720 -0.9207 -0.7229 0.1395 0.8737 0.8046
-0.1510 0.1163 0.2766 0.1826 -0.0793 -0.2683 -0.2107 0.0407 0.2546 0.2344
0.3550 -0.2734 -0.6504 -0.4294 0.1864 0.6308 0.4953 -0.0956 -0.5986 -0.5513
0.5346 -0.4117 -0.9795 -0.6467 0.2806 0.9500 0.7459 -0.1440 -0.9014 -0.8301
0.2227 -0.1715 -0.4080 -0.2694 0.1169 0.3957 0.3107 -0.0600 -0.3755 -0.3458
-0.2939 0.2264 0.5386 0.3556 -0.1543 -0.5224 -0.4101 0.0792 0.4957 0.4565

which i hope is correct?

But i have no idea how to do the same thing with a loop-free script.

 

[DryRun]

Here is my attempt for part 2 but i think there is a shorter version:

k=sin(1:1:10);
m = cos(1:1:10);
r1=k(1).*m
r2=k(2).*m
r3=k(3).*m
r4=k(4).*m
r5=k(5).*m
r6=k(6).*m
r7=k(7).*m
r8=k(8).*m
r9=k(9).*m
r10=k(10).*m
A=[r1;r2;r3;r4;r5;r6;r7;r8;r9;r10]

 

[NemoReally]

I have attached the problem to this post. My attempt at the first part (i used 'm' instead of 'l' as it's less confusing, since the latter resembles the digit '1') and here is my script:

A = zeros(10);
for k=1:10 
  for  m=1:10
    A(k,m)  =  sin(k)*cos(m); 
  end 
end 
 
A

The answer:

A =
0.4546 ...

which i hope is correct?

Well, they seem to agree with my Mathcad-generated results, if that's of some reassurance.

But i have no idea how to do the same thing with a loop-free script.

Try something like:
k = 1:10;
A = sin(k)' * cos(k);

note the single quote mark, the transpose operator. The creates a vector, k, with values 1 to 10, the sin and cos functions operate over vectors, resulting in 2 vectors, the sin is transposed and the resulting matrix multiplication is equivalent to the first loop method.

You can use just k as the number of k and m elements are the same.

 

[NemoReally]

Forgot to add, image showing proof of pudding in Mathcad.

 

[DryRun]

Try something like:
k = 1:10;
A = sin(k)' * cos(k);

note the single quote mark, the transpose operator. The creates a vector, k, with values 1 to 10, the sin and cos functions operate over vectors, resulting in 2 vectors, the sin is transposed and the resulting matrix multiplication is equivalent to the first loop method.

You can use just k as the number of k and m elements are the same.

That's an ingenious way of solving it, as a 10x1 matrix multiply by another 1x10 matrix gives the required 10x10 matrix!

Thank you very much, NemoReally.

 

[NemoReally]

That's an ingenious way of solving it, as a 10x1 matrix multiply by another 1x10 matrix gives the required 10x10 matrix!

It is, isn't it.


The important thing about languages such as Mathcad, Mathematica, Maple, Matlab and J is learning to get one's head out of the detailed programming gutter and looking up into the higher level world of mathematics, particularly arrays and functions. There are still occasions when a bit of close quarters coding is required, but its far better to be able to think about multiplying two matrices at company commander level rather than march them around the binary parade square yourself. (I now it declare it International Mixed Metaphor Day.) Imagine you're writing the problem down on a whiteboard and then see what methods exist to support what you've written.

Thank you very much, NemoReally.

No worries.