Find the matrix exponential $e^{At}$ for $$A=\begin{pmatrix} 1& 0\\ 3& 1\\ \end{pmatrix}$$
I split this into the matrices
$$B=\begin{pmatrix} 1& 0\\ 0& 1\\ \end{pmatrix}$$
and
$$C=\begin{pmatrix} 0& 0\\ 3& 0\\ \end{pmatrix}$$
And solved $e^{Bt}=\begin{pmatrix} e^t& 0\\ 0& e^t\\ \end{pmatrix}$ and $e^{Ct}=\begin{pmatrix} 0& 0\\ e^{3t}& 0\\ \end{pmatrix}$
I simply used power series for the exponential since $C^n=\begin{pmatrix} 0& 0\\ 3^n& 0\\ \end{pmatrix}$ and $B^n=\begin{pmatrix} 1^n& 0\\ 0& 1^n\\ \end{pmatrix}$
Then I computed $e^{Bt}e^{Ct}=e^{At}$
I got $e^{At}=\begin{pmatrix} 0& 0\\ e^{4t}& 0\\ \end{pmatrix}$
But when I check this with a exponential calculator, it gives a solution $\begin{pmatrix} e^t& 0\\ 3e^t t& e^t\\ \end{pmatrix}$
What have I done wrong?
Best Answer
You have to do directly from the matrix $A$, observe that $$ \big(At\big)^n=\begin{pmatrix} t^n & 0\\ 3nt^n & t^n \end{pmatrix} $$ then $$ e^{At}=\sum_{n\geq1}\frac{\big(At\big)^n}{n!}=\begin{pmatrix} \sum_{n\geq 1}\frac{t^n}{n!} & 0\\ \sum_{n\geq 1}\frac{3nt^n}{n!} & \sum_{n\geq 1}\frac{t^n}{n!} \end{pmatrix}=\begin{pmatrix} e^t & 0 \\ 3t\sum_{n\geq 1}\frac{t^{n-1}}{(n-1)!} & e^t \end{pmatrix}=\begin{pmatrix} e^t & 0 \\ 3te^t & e^t \end{pmatrix} $$