Inverse of higher order matrix

Hello there,

I am trying to get inverse of a 80,000 order matrix but the calculation is taking too much time. Is there any fastest way or method to calculate the inverse of a matrix?

[Stephen23]

Why do you need the inverse of the matrix? What will you do with the inverse?

Read this: https://www.mathworks.com/help/matlab/ref/inv.html#bu8tv8c-6

[raina RAJ]

It is needed in some of my calculations and there is no other way out. I have to get the inverse. I am using the backslash command but still the system gets hang everytime.

[Stephen23]

"It is needed in some of my calculations and there is no other way out. I have to get the inverse. I am using the backslash command but still the system gets hang everytime."

Help us to help you, don't make it hard for us! There are many regular users on this forum who have many many years of experience working with matrices and linear algebra, you might as well take advantage of this amazing resource. If you give us absolutely no information at all, then we can't help you figure out if it is possible to solve your problem efficiently.

[raina RAJ]

I have system of equations xQ = 0 and xe =1 where 'e' is a column vector of appropriate dimension, x is a row vector of size (1,M) and Q matix has size (M,M). I need to find out the solution x here. In order to do that, I need to solve the system. I am solving this using backslach operator. But the size of the matrix is big (80000 types), so the system gets hang everytime. The system has 8 GB RAM and 64 bit operating system. Please help me out.

[Matt J]

Is the matrix Q sparse? Are you using Matlab's sparse data format to store Q?

Does e contain arbitrary values or is it e=ones(M,1)?

[raina RAJ]

Yes matrix Q is stored in sparse data format and e is ones(M,1).

Answer 1

[Matt J]

I have system of equations xQ = 0 and xe =1 where 'e' is a column vector of appropriate dimension, x is a row vector of size (1,M) and Q matix has size (M,M).

If Q is sparse, you can do,

x=eigs(Q.',1,'sm');

x=x/sum(x);

Answer 2

[Christine Tobler]

Another way of looking at the equations x*Q = 0 and x e = 1 is as an equation system x * [Q e] = [0 1], which you could solve using

A = [Q, ones(n, 1)]';
b = [zeros(n, 1); 1];
x = A \ b;

or if this is too slow, you could try an iterative algorithm:

x = lsqr(A, b);