How to find inverse of large size matrix.?

I have a program in which I have need to find inverse of large size square matrix..e.g. [m n]=[800 800 ]

It takes longer to solve the program. I want to more effective and time saving way to solve that.

[Jan]

Without knowing your code and what "longer" exactly means, how can we help you?

Answer 1

[John D'Errico]

800x800 is not even remotely large, IMO.

Everyone assumes that computers are infinitely fast, infinitely powerful, and that there are always tricks one can do to speed up any computation.

Computers are only as fast as they are. Computing the inverse of a matrix (something you almost never really need to do anyway, IF you know what you are doing) is a computation that takes a known number of adds, multiplies, etc. Your computer can only do those operations as fast as it can.

So, IF you truly need to compute an inverse and the time below is not fast enough for you, then...

A = rand(800);
timeit(@() inv(A))
ans =
     0.054773

So it looks like my computer is a little faster than Jan's.

Regardless, if this is not fast enough, then you can get a faster computer. That is the best way to solve your problem. As you see, my computer solved the problem roughly 3 times as fast as did Jan's.

Next, you MIGHT be able to do things with sparse matrices, IF you know what you are doing, since you will need to not compute a direct inverse, AND IF your matrix is truly a sparse matrix. Many people think their matrices are sparse when they are not even close to sparse. The inverse of a sparse matrix will not in general be sparse, and so it may actually be slower to compute. So you will then need to reformulate the problem to avoid computing an inverse.

Of course, if you are able to reformulate the problem, then some things can yield speedups without too much effort. For example, an alternative to computing an inverse, is

B = eye(800);
timeit(@() A\B)
ans =
   0.049547

As you can see here it gained about 10%, simply by using backslash instead of inv.

Other ideas might be to try moving the problem into your GPU. But that will require you have the parallel computing toolbox, AND learn how to use it effectively, and even then, I have no idea if you will see any real gain.

So the best way to see a speed increase is to get a faster computer. Or, get a cup of coffee and read a good book while you wait. If you want a better answer, it would help if we knew more about the matrices you are trying to invert, and specifically what you are doing with that inverse.

[Walter Roberson]

On my 2012 Mac, the time for the inv() is roughly 0.023, more than twice as fast as John's. On my system, the \ solution is slower, about 0.026

[John D'Errico]

Not sure when my Mac dates to, but maybe a year or so older then Walter. Mine was a reasonably fast machine at the time because I was doing some consulting in those days, but not fully tripped out even then.

It does imply that a new system, with an up to date, decent processor in it should give you the best gain in speed.

[Walter Roberson]

"About this Mac" will tell you the year.

Mine is a 2.6 GHz Intel i7 with 4 physical cores. If yours is older is is probably not an i7 and possibly it does not have 4 physical cores. inv() scales well with number of cores in that range.

[John D'Errico]

Nope. Identical specs on mine. 4 cores. i7, 2.8 GHz.

[Walter Roberson]

John, is it possible that your system is the late 2009 27" iMac? That was the Core i7 (860) ("Wolfdale, Lynnfield") architecture. The alternative is that it is the mid 2011 21.5" model with Core i7 (2600S) ("Sandy Bridge") -- those are the only iMac that had 2.8 GHz i7 available.

My MacBook Pro is Intel Core i7 (3615QM, 3720QM, 3820QM) ("Ivy Bridge"). If your system was 2 architecture generations behind then the major speed difference would be understandable.

In the time since I got my Sandy Bridge machine, they have gone through Haswell, Broadwell, and Skylake so improved performance since mine would be expected.

[Rik]

Flag by Adam Ward:

This is quite possibly one of the most obnoxious answers I have ever read on mathworks. Thanks.

It is easy to misread the tone if you only have the text. In this answer John takes time to explain the situation and the possible solutions in some detail. Even if this was intended to be rude/obnoxious, this isn't the most obnoxious by a long shot. You understandably don't see those often, because they get edited down to remove the rudeness.

I hope you have this thread in your activity feed so you actually see this comment. Do you have any suggestions for edits?

[Bruno Luong]

Hmmm, inv is faster in my 2018 Asus Laptop (R2019a)

>> A = rand(800);
>> timeit(@() inv(A))

ans =

    0.0172

>> B = eye(800);
>> timeit(@() A\B)

ans =

    0.0248

[Rik]

time_list=zeros(20,2);
for n=1:size(time_list,1)
    A = rand(800);
    B = eye(800);
    time_list(n,1)=timeit(@() inv(A));
    time_list(n,2)=timeit(@() A\B);
end
mean(time_list)
%         inv        \
%R2019a: 0.0268    0.0331
%R2015a: 0.0297    0.0361

So maybe John's initial result was a fluke? max(time_list(:,1))>min(time_list(:,2)) returns false for R2019a, but true for R2015a. I don't have 2017a installed which was presumably the release used.

(W10x64, i7-4710MQ@2.5GHz)

[Bruno Luong]

"timeit calls the specified function multiple times, and computes the median of the measurements."

It seems redundant to loop on it

[Rik]

True, but there is still some spread, and since A is different every iteration, there is an extra source of variation. You could time @() inv(rand(800)) instead, but then you no longer have an idea about the spread, which could explain a fluke result.

Answer 2

[Jan]

The explicite calculation of inverse is seldom necessary. Are you really sure you need it?

x = rand(800);
tic;
y = inv(x);
toc
% Elapsed time is 0.187157 seconds.

Do you need it much faster?