Inverse matrix in Matlab very different to Excel

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Rosie G
Rosie G 2019-12-10
评论： Fabio Freschi 2019-12-10
Replicating calculations carried out in Excel in Matlab in order to learn. Inverting the matrix below gave a very different answer in Matlab to excel.
Original Matrix:
1322252 0 2010 -1006
0 1322252 1006 2010
2010 1006 4 0
-1006 2010 0 4
Inverted Matrix using inv(matrix)
1.68867574048431e-05 5.0821976835258e-21 -0.00848559559593365 0.00424701948731803
5.0821976835258e-21 1.68867574048431e-05 -0.00424701948731803 -0.00848559559593364
-0.00848559559593365 -0.00424701948731803 5.58213718801715 -1.28379263845227e-15
0.00424701948731803 -0.00848559559593364 -1.28379263845227e-15 5.58213718801714
Inverted Matrix using Excel
1.68868e-05 -1.34343e-20 -0.008485596 0.004247019
2.82121e-20 1.68868e-05 -0.004247019 -0.008485596
-0.008485596 -0.004247019 5.582137188 -6.91897e-15
0.004247019 -0.008485596 6.91897e-15 5.582137188
Inverting the below matrix did work as expected (same answer in Excel and Matlab)
419 163 1 0
-163 419 0 1
586 340 1 0
-340 586 0 1
  1 个评论
Jeremy
Jeremy 2019-12-10
This is probably a case of different methods and/or numerical tolerances. The results are approximately the same, 5.0821976835258e-21 and -1.34343e-20 are both essentially zero

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Fabio Freschi
Fabio Freschi 2019-12-10
The answer is not very different. As Jeremy commented, the results are similar a part from numerical tolerances. You can verify yourself comparing the matrices. In the following I load your matrices (note that excel data have less digits):
A1 = [ 1.68867574048431e-05 5.0821976835258e-21 -0.00848559559593365 0.00424701948731803
5.0821976835258e-21 1.68867574048431e-05 -0.00424701948731803 -0.00848559559593364
-0.00848559559593365 -0.00424701948731803 5.58213718801715 -1.28379263845227e-15
0.00424701948731803 -0.00848559559593364 -1.28379263845227e-15 5.58213718801714];
A2 = [ 1.68868e-05 -1.34343e-20 -0.008485596 0.004247019
2.82121e-20 1.68868e-05 -0.004247019 -0.008485596
-0.008485596 -0.004247019 5.582137188 -6.91897e-15
0.004247019 -0.008485596 6.91897e-15 5.582137188];
Then you can compare the two matrices entry-by-entry:
A1-A2
ans =
-4.2595e-11 1.8516e-20 4.0407e-10 4.8732e-10
-2.3130e-20 -4.2595e-11 -4.8732e-10 4.0407e-10
4.0407e-10 -4.8732e-10 1.7150e-11 5.6352e-15
4.8732e-10 4.0407e-10 -8.2028e-15 1.7140e-11
And you can see the errors are "small" in absolute value (the largest difference is at the 10th digit)
As a global index you can use the (relative) norm of the matrices
norm(A1-A2)/norm(A1)
ans =
1.1581e-10
  2 个评论
Rosie G
Rosie G 2019-12-10
Thank you so much for your fast reply, and I feel very foolish now!
Fabio Freschi
Fabio Freschi 2019-12-10
No problem, sometimes obvious things slip under the radar!

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