Doubt in the defn of exponential operator. definition
Let $A$ be an $n\times n$ matrix. Then for $t\in \mathbb R$,
$$e^{At}=\sum_{k=0}^\infty \frac{A^kt^k}{k!}\tag{1}$$
But in this definition, What they are meaning by the term $A^kt^k$, If I give this matrix $$A=\begin{pmatrix}1&-1\\2&3\end{pmatrix},$$
what can we say about the $(1)$, means  How it will be?
 A: To do exponentiation of matrices, the easiest way is to diagonalize the exponent first, 
$U^{\dagger} AtU =  D = \mbox{diag}\{D_1, D_2,...,D_n\}  $ with eigen values $D_1,D_2,...,D_n$
since
$ U^{\dagger}exp(At) U = exp(D) = \mbox{diag}\{e^{D_1}, e^{D_2},...,e^{D_n} \} $
Then 
$$exp(At) = U exp(D)U^{\dagger} $$
For you matrix, you will get, 
$$ e^{At} = \left(
\begin{array}{cc}
 e^{2 t} \cos (t)-e^{2 t} \sin (t) & -e^{2 t} \sin (t) \\
 2 e^{2 t} \sin (t) & e^{2 t} \cos (t)+e^{2 t} \sin (t) \\
\end{array}
\right) .$$
A: One way to handle this kind of problem is to find the Jordan Normal Form of $A$:
$$
A=\begin{bmatrix}
-1-i&-1+i\\
2&2
\end{bmatrix}
\begin{bmatrix}
2-i&0\\
0&2+i
\end{bmatrix}
\begin{bmatrix}
-1-i&-1+i\\
2&2
\end{bmatrix}^{-1}
$$
Taking the exponential of $A$ is now pretty simple:
$$
\begin{align}
\exp(At)&=
\begin{bmatrix}
-1-i&-1+i\\
2&2
\end{bmatrix}
\begin{bmatrix}
e^{2t}e^{-it}&0\\
0&e^{2t}e^{it}
\end{bmatrix}
\begin{bmatrix}
-1-i&-1+i\\
2&2
\end{bmatrix}^{-1}\\
&=\begin{bmatrix}
-1-i&-1+i\\
2&2
\end{bmatrix}
\begin{bmatrix}
e^{2t}e^{-it}&0\\
0&e^{2t}e^{it}
\end{bmatrix}
\begin{bmatrix}
2i&1+i\\
-2i&1-i
\end{bmatrix}\\
&=e^{2t}\begin{bmatrix}
\cos(t)-\sin(t)&-\sin(t)\\
2\sin(t)&\cos(t)+\sin(t)
\end{bmatrix}
\end{align}
$$
A: Answering your question $(1)$ will be
$$ e^{At} = \left[ \begin {array}{cc} {{\rm e}^{2\,t}}(\cos \left( t \right)- \sin \left( t \right)) &-{{\rm e}^{2\,t}}\sin \left( t
 \right) \\ 2\,{{\rm e}^{2\,t}}\sin \left( t
 \right) & {{\rm e}^{2\,t}}(\cos \left( t \right) +\sin\left( t \right)) \end {array} \right] .$$
