Computing the inverse of a matrix without using the 'backslash' command

So I am trying to compute the inverse of a matrix, and multiply it by another matrix. When I evaluate my code I get two answers on the order of 10^20. It is because the matrix is singular, and cannot be easily inverted. Is there a way my code can be evaluated to obtain correct values?

function solver=partone(Aee, Aet, Ate, Att, De, Dt)

A=[Aee Aet;Ate Att];
d=[De;Dt];
I=eye(2);

solver=(I-A)\d;

In my code I am using A=[.5 .3;.3 .82] and d=[110000;-40000]

[Sebastian Holmqvist]

"A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if and only if its determinant is 0."

So is it singular or not? You can not invert a singular matrix since there's no inverse.

[Nathaniel]

The determinate of the matrix is zero so it would be singular. It is for a school project and there is a way to get an answer. I have been told that one of the solutions will approach infinity and the other will approach zero. I am unsure how to figure this out myself.

Answer 1

[Puneet Rana]

You can use the Moore-Penrose pseudoinverse as follows:

solver=pinv(I-A)*d

[Nathaniel]

Thanks!

[Richard Brown]

You do realise that this is still not a "solution" to your equations though, right? They have no solution, because they are inconsistent.

Because they are inconsistent, the best you can do is find an x that minimises the (Euclidean norm of) the residual (I - A)*x - d. Because your system has rank 1, there is a 1D subspace of R^2 that has this property. The pinv solution presented here finds the x of minimum norm from this set of minimisers.