number divided by vector

Hi everyone~ I divided a number by a vector and forgot to use ./ Why it didn't report error and gave me some result? What's the meaning of this calculation? Thank you!

t = (1: 0.5: 50)';
Carnot = 1-295/(500*exp(-0.02*t)+300);

Answer 1

[yonatan gerufi]

This is a tricky thing.

a. the command:

 10/ [10,20]'

is equal (from matlab point of view) to:

[10;0]/ [10;20]

but the results will be just the non zeros line.

b. the command:

 10/ [50,20]' (first element in denominator is larger)

is equal (from matlab point of view) to:

[0;10]/ [10;20]

but the results will be just the non zeros line.

[Yuji Zhang]

Thanks Yonatan! I did some test and the results seem to be different...

clear;

t = (1: 3)';
y = exp(-0.02*t)+1;
c1 = 1/y;
c2 = [1 0 0]'/y;

Also do you know why 1/[1 2] yields error and 1/[1 2]' is legal? Let me know please. Thanks!

Answer 2

[yonatan gerufi]

that's a good question, i believe both needs to be illegal.

how do you suggest matlab will understand the command:

1/[1,2]' ?
[1;0]/[1;2] or [1;1]/[1;2] or perhaps [0;1]/[1;2] ?

there is lack of clarity.

for better understanding i recommend reading on matrix divide. see for example: link 1,

if the previous answer answers your question you can do "accept answer" so people will know.

good luck!

Answer 3

[Yuji Zhang]

I think I figured out myself.

Division is defined by multiplication. For any A, B, and C, A/B = C means A = B*C.

Note B*C = C*B works for numbers, but not for vectors and matrices.

Think of x = 1/[1 2]'. What is x? It is something that satisfies x*[1 2]' = 1.

Without knowing the specific value, we know x is a row vector of length = 2, because only then x*[1 2]' will be a real number.

In fact

>> 1/[1 2]'

ans =

           0    0.5000

>> [0 0.5]*[1 2]'

ans =

       1

On the other hand

>> 1/[1 2]
Error using  / 
Matrix dimensions must agree.

Think of x = 1/[1 2], then x*[1 2] = 1, and there doesn't exist a solution. So it is illegal.

Note 1/matrix is called the inverse: 1/M = inv(M) There is a pseudo inverse pinv() that yields a best estimation for matrices that don't have an inverse, and also for vectors.

>> 1/[1 2]
Error using  / 
Matrix dimensions must agree.

>> pinv([1 2])

ans =

      0.2000
      0.4000

>> pinv([1 2]) * [1 2]

ans =

      0.2000    0.4000
      0.4000    0.8000

The answer is not 1. I think that's the best estimation by some algorithm.

Answer 4

[Shoaib Waqar]

you can try

1./matrix