# Consequences on a system of ODEs after performing operations

• I
Hi, I have derived a matrix from a system of ODE, and the matrix looked pretty bad at first. Then recently, I tried the Gauss elimination, followed by the exponential application on the matrix (e^[A]) and after another Gauss elimination, it turned "down" to the Identity matrix. This is awfully simple, and I may wonder if the system at hand is either completely redundant, or if the procedure of using Gauss elimination on a matrix for a system of ODEs, followed by e^A and then another Gauss elimination is a reasonable and valid approach to solve ODE systems, and does not "make critical information of the ODE vanish along the way"?

Thanks!

Homework Helper
Not clear what you are doing. What is in the first matrix ? What does Gauss elimination give you when you do that ? Can you give an example ?

Not clear what you are doing. What is in the first matrix ? What does Gauss elimination give you when you do that ? Can you give an example ?

The first matrix is:

A = \begin{bmatrix}
1.0545718E-68 & 1.1271866E-(61i) \\
1 & -1\\
\end{bmatrix}

the first operation in gauss elimination with rref[A] in matlab

that gives:

B = \begin{bmatrix}
1 & -1 \\
0 & 0\\
\end{bmatrix}

the exponential of B gives then:

C= \begin{bmatrix}
2.7183 & 0.3679 \\
1.0000 & 1.0000 \\
\end{bmatrix}

finally gauss elim of C with rref[C] gives the identity matrix

Does that make sense? I suspect MATLAB is rounding exponentially small values to 0, and this is not a correct result.

Thanks

Mentor
2022 Award
Due to the very small first row, the matrix has been interpreted as ##A=\begin{bmatrix}0&0\\1&-1\end{bmatrix}## by the program, which is a singular matrix. It results in ##\exp(A)=\begin{bmatrix}1&0\\1+e^{-1}&e^{-1}\end{bmatrix}##.

Now that you exchanged the rows of ##A## for whatever reason, you have ##B=\begin{bmatrix}1&-1\\0&0\end{bmatrix}## instead, which leads to ##\exp(B)=\begin{bmatrix}e&e^{-1}\\0&1\end{bmatrix}##, if I made no mistakes by exponentiation.

So what do you want to do or say? As it is a numerical error, which led to that result, almost zero instead of equal to zero, we have algorithmic stability as one topic, or the general case of a Lie Group and its tangent space: ##\mathfrak{g} \stackrel{\exp}{\longrightarrow} G## as another.

• SeM
Thanks, I prefer to rectify the numerical error! Cheers