Compute $e^A$ where $A$ is a given $2\times 2$ matrix Compute $e^A$ where $A=\begin{pmatrix} 1 &0\\ 5 & 1\end{pmatrix}$
definition
Let $A$ be an $n\times n$ matrix. Then for $t\in \mathbb R$,
$$e^{At}=\sum_{k=0}^\infty \frac{A^kt^k}{k!}\tag{1}$$
 A: Or, write
$A = I + N \tag{1}$
with
$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \tag{2}$
and
$N = \begin{bmatrix} 0 & 0 \\ 5 & 0 \end{bmatrix}.  \tag{3}$
Note that
$IN = NI = N, \tag{4}$
that is, $N$ and $I$ commute, $[N, I] = 0$, and apply the well-known result that for commuting matrices $B$ and $C$ we have $e^{B + C} = e^B e^C$, an exposition of which may be found here:  $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$,
and write
$e^A = e^I e^N. \tag{6}$
Now using the matrix power series definition of $e^X$,
$e^X = \sum_0^\infty \dfrac{X^n}{n!}, \tag{7}$
it is easily seen that
$e^I = \sum_0^\infty (\dfrac{1}{n!} I) = (\sum_0^\infty \dfrac {1}{n!}) I = e I, \tag{8}$
since $I^n = I$ for all $n$, while we easily calculate
$N^2 = 0, \tag{9}$
so with $X = N$ (7) becomes
$e^N = I + N = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}. \tag{10}$
Then
$e^A = e^{I + N} = e^I e^N$
$ = e^I(I + N) = e I(I + N) = e (I + N) , \tag{11}$
and finally, again with the aid of (10),
$e^A = \begin{bmatrix} e & 0 \\ 5e & e \end{bmatrix}.  \tag{12}$
A: i will turn it into $$\frac{dx}{dt} = x, \frac{dy}{dt} = 5x + y$$ which has solution $$x = Ae^t, \frac{d}{dt}\left(ye^{-t}\right) = 5A \to y = Be^t + 5At e^t $$
we can write $$\pmatrix{x\\y} = \pmatrix{e^t&0\\5te^t&e^t}\pmatrix{x(0)\\y(0)} $$ by a property of $e^{At}$ this implies $$ e^{tA} = \pmatrix{e^t&0\\5te^t&e^t},\, e^{A} = \pmatrix{e&0\\5e&e} $$
A: Another way of proving this...
we know from  the property of $e$, that if $S,T$ are linear operators and $ST=TS$ then 
$$e^{S+T}=e^S.e^T$$
Letting $S=I$ and $T=\begin{pmatrix}0&0\\5&0\end{pmatrix}$
Then $$e^{S+T}=e^S.e^T=e^I.e^{\begin{pmatrix}0&0\\5&0\end{pmatrix}}=e(I+T)=e\begin{pmatrix}1&0\\5&1\end{pmatrix}$$
Since, $T^2=0=T^3=\ldots$
