# number divided by vector

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Yuji Zhang on 9 Nov 2014
Answered: Shoaib Waqar on 22 Jan 2020
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);
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### Answers (4)

yonatan gerufi on 9 Nov 2014
Edited: yonatan gerufi on 9 Nov 2014
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.
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Yuji Zhang on 9 Nov 2014
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!

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yonatan gerufi on 9 Nov 2014
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!
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Yuji Zhang on 9 Nov 2014
Edited: Yuji Zhang on 9 Nov 2014
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.
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Shoaib Waqar on 22 Jan 2020
you can try
1./matrix
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