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When I try to calculate a matrix inverse using Matlab's inv() operation:

A = rand(10,10);
b = rand(10,1);

C = inv(A);
D = C*b;

I get the following warning at the last row: INV is slow and inaccurate. Use A\b for INV(A)*b , and b/A for b*INV(A).

I can change the code above into:

A = rand(10,10);
b = rand(10,1);

C = inv(A);
D = A\b;

Now I don't get the warning, but I don't think this solution is better.

Note: I need to store both the inverse of matrix A as well as inv(A)*c. Also, in my real file the size of matrix A can be 5000 x 5000 or even bigger.

Are there any better solutions in terms of efficiency and accuracy or is the first method fine?

Amro
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Pietair
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    Related: [Why is Matlab's inv slow and inaccurate?](http://stackoverflow.com/questions/1419580/why-is-matlabs-inv-slow-and-inaccurate) – HebeleHododo May 14 '13 at 14:31

3 Answers3

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If you absolutely need the inverse later on, then you have to compute it. If you can use the backslash operator (\) instead of an inverse again later on, I would stay away from the inverse and listen to MATLAB's suggestion. For numerical reasons, it is always better to use a slash operator when you can, so the second approach is better even though it is slower.

Marc Claesen
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  • I agree. Also, `tic`-`toc` showed that the second approach is slower: `0.0851` against `0.1284` – rascob May 14 '13 at 14:32
  • @rascob The second option is slower because it is doing more work (inv and \ ), but \ gives a more accurate answer. – shoelzer May 14 '13 at 18:07
  • @shoelzer Sure, wasn't surprised by the result :) Just wanted to show that, since inv(A) will be needed anyhow, the first approach is indeed faster. – rascob May 14 '13 at 18:35
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    @rascob Right. But even if `inv(A)` is needed at some point, it's not needed here. I think it's always better use `A\b` if possible. – shoelzer May 14 '13 at 18:44
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You should listen to Matlab and use the second option. inv(A)*b and A\b are computed with different algorithms under the hood, and \ is indeed more accurate.

The documentation for inv states:

In practice, it is seldom necessary to form the explicit inverse of a matrix. A frequent misuse of inv arises when solving the system of linear equations Ax = b. One way to solve this is with x = inv(A)*b. A better way, from both an execution time and numerical accuracy standpoint, is to use the matrix division operator x = A\b. This produces the solution using Gaussian elimination, without forming the inverse. See mldivide () for further information.

shoelzer
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  • I'v same Problem, I used both `/` and `\` but couldn't resolve my case, not getting accurate value. Also I only want first index only, i.e `ans(1)` . My code: `inv(-[40 200;1 -1])*(0.9*(2-200))` –  Apr 27 '14 at 16:42
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Some additional information:

If you are to calculate

Ax = b

For many different b's, but with a constant A, you might want to pre-factorize A. That is:

[L U P] = lu(A);
x = (U \ (L \ ( P * b)));

Don't know about other fields, but this occurs frequently in power system engineering at least.

Stewie Griffin
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