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The fix(x) command does not work

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Oskar
Oskar on 29 Nov 2020
Answered: Steven Lord on 29 Nov 2020
Why does the one not become a one when i fix it. I have tried everything but cant get it to work!
A=[4 2 3 1; 2 5 6 2; 0 0 1 3; -1 -2 9 8]
B=inv(A);
C=A*B;
C = fix(C)
D=B*A;
D = fix(D)
C==D

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John D'Errico
John D'Errico on 29 Nov 2020
Please don't post a PICTURE of code. That makes it impossible for people to copy what you did, and then use it as an example to help you. Is there a good reason why you WANT to make it more difficult for someone to help you???????

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Accepted Answer

John D'Errico
John D'Errico on 29 Nov 2020
Edited: John D'Errico on 29 Nov 2020
Fix does work. You need to understand what you asked it to do.
A = [4 3 2 1;2 5 6 2;0 0 1 3;-1 -2 9 8];
format long g
B = inv(A)
B =
0.361823361823362 -0.190883190883191 -0.168091168091168 0.0655270655270655
-0.153846153846154 0.230769230769231 0.307692307692308 -0.153846153846154
0.00854700854700855 0.0427350427350427 -0.350427350427351 0.11965811965812
-0.00284900284900285 -0.0142450142450142 0.45014245014245 -0.0398860398860399
Do you recognize the matrix inverse is NOT composed of integer values?
C = A*B
C =
1 8.15320033709099e-17 5.55111512312578e-17 -3.46944695195361e-17
-1.12757025938492e-16 1 3.33066907387547e-16 9.71445146547012e-17
-3.46944695195361e-18 0 1 -1.38777878078145e-17
7.63278329429795e-17 9.71445146547012e-17 -4.44089209850063e-16 1
The resulting product is NOT identically an identity matrix. That is exactly as expected. The deviations from an identity matrix are tiny, errors in the least significant bits.
fix(C)
ans =
0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
C(1,1)
ans =
1
>> C(1,1) == 1
ans =
logical
0
>> C(1,1) - 1
ans =
-1.11022302462516e-16
So C(1,1) is not exactly 1. In fact, it is just a tiny bit less than 1. Close to 1 is not exactly 1. For example:
fix(1 - eps)
ans =
0
Is the result an identity matrix? It should be effectively so, if A was of full rank.
rank(A)
ans =
4
norm(A*B - eye(4))
ans =
6.74932945500581e-16
Again, all completely expected.
Since A is composed of integers, and A is of full rank, you could do this:
Bs = inv(sym(A))
Bs =
[127/351, -67/351, -59/351, 23/351]
[ -2/13, 3/13, 4/13, -2/13]
[ 1/117, 5/117, -41/117, 14/117]
[ -1/351, -5/351, 158/351, -14/351]
A*Bs
ans =
[1, 0, 0, 0]
[0, 1, 0, 0]
[0, 0, 1, 0]
[0, 0, 0, 1]
The result is now an EXACT identity matrix, because Bs is also exact, which was possible because A was an integer matrix.
Note that symbolic inverses on large matrices will typically be massively computationally intensive.

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More Answers (1)

Steven Lord
Steven Lord on 29 Nov 2020
Just because a number is displayed as 1 in the default display format does not mean that the stored value is exactly 1. It's close but not exactly equal to 1. It's ever so slightly less than 1.
A=[4 2 3 1; 2 5 6 2; 0 0 1 3; -1 -2 9 8];
B=inv(A);
C=A*B;
1 - C(2, 2)
ans = 1.1102e-16

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