The fix(x) command does not work

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

[John D'Errico]

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???????

Answer 1

[John D'Errico]

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.

Answer 2

[Steven Lord]

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