inverse of a matrix wrong??

Hello community,

I've just written this code:

close all
clear all
clc

s = tf('s');
A = [0 , 1 ; -2 , -3]
B = [1 ; 0]
I = [1 , 0 ; 0 , 1]

L = s*I-A;

Linv = inv(L)

and it's not equal to my counts that I've done on the paper; infact the L matrix should be

s+3/(s^2+3*s+2) 1/(s^2+3*s+2)

-2/(s^2+3*s+2) s/(s^2+3*s+2)

while in matlab the last fraction ( row 2 , column 2) is: (s + 8.438e-15)/(s^2+3*s+2)

Answer 1

[Walter Roberson]

It is not wrong, but it is subject to round-off error; it also is not being simplified.

[Giovanni Salviati]

So the result in Matlab is more accurate? And I can't do nothing to make matlab outputting the result I can calculate on my own? Because I don't know if in the exam the lecturer accepts this kind of result (matlab result)

[Walter Roberson]

To calculate it without round-off error you will probably need to switch to Symbolic Toolbox

syms S
inv([S, -1;2, S+3])

[Giovanni Salviati]

Sorry but I didn't get it. How does it work?

[Giovanni Salviati]

Ok solved it.

I've just written "syms s", instead of "s = tf('s')"

but i don't get it why with the second function it doesn't work

[Walter Roberson]

The underlying code for tf and ss systems works with pure numeric matrices, and those have round off errors in calculations of matrix inverse. Matrix inverse numeric calculations are not done according to the formal definition of matrix inverse, because those calculations can take a long time.