Exponentiation with Matrices How would exponentiation be done with matrices?
For example:
$$ \left[\begin{array}{cc}1&2\\8&7\end{array}\right]^{\left[\begin{array}{cc}7&4\\2&9\end{array}\right]}$$
Where both base and exponent are matrices.
$$ \left[\begin{array}{cc}1&2\\8&7\end{array}\right]^6$$
Where the base is a matrix, and the exponent is not.
$$6^{\left[\begin{array}{cc}7&4\\2&9\end{array}\right]}$$
Where the base is not, and the exponent is a matrix.

Wolfram Alpha doesn't recognize it, which makes me wonder if it is even possible.
 A: Using abel and Peter's comments, I was able to solve my first question (matrix exponentiation with two matrices).
Please let me know if I am incorrect in any of my logic.
$$\text{Using a random example:}
\\
a = \begin{bmatrix}
1 & 2\\ 
8 & 7
\end{bmatrix}
\\
b = \begin{bmatrix}
7 & 4\\
2 & 9
\end{bmatrix}
\\
a^{b} = e^{b\ln(a)}
\\
\ln(a) = \begin{bmatrix}\ln(1)&\ln(2)\\\ln(8)&\ln(7)\end{bmatrix}
\\
\\
a^b = e^{\begin{bmatrix}7&4\\2&9\end{bmatrix}\times\begin{bmatrix}\ln(1)&\ln(2)\\\ln(8)&\ln(7)\end{bmatrix}}
\\
a^b = \begin{bmatrix}
7 & 16\\
256 & 4782969
\end{bmatrix}
$$
Wolfram Alpha corroborates the final exponential arithmetic, $e^{\begin{bmatrix}7&4\\2&9\end{bmatrix}\times\begin{bmatrix}\ln(1)&\ln(2)\\\ln(8)&\ln(7)\end{bmatrix}}$, link here.
A: *

*Concerning the logarithm of a matrix
The logarithm (understood as the inverse of exp) of a matrix is not the log of each component:
example
$$
   \exp \left( {\left( {\begin{array}{*{20}c}    0 & 0 & 0  \\    1 & 0
   & 0  \\    0 & 1 & 0  \\  \end{array} } \right)} \right) = \left(
   {\begin{array}{*{20}c}    0 & 0 & 0  \\    1 & 0 & 0  \\    {1/2} & 1
   & 0  \\  \end{array} } \right)\quad \quad \exp \left( {\left(
   {\begin{array}{*{20}c}    1 & 0 & 0  \\    1 & 1 & 0  \\    0 & 1 & 1
   \\  \end{array} } \right)} \right) = \left( {\begin{array}{*{20}c}   
   e & 0 & 0  \\    e & e & 0  \\    {1/2e} & e & e  \\  \end{array} }
   \right)
   $$
To my knowledge, it can instead be defined through the  Taylor
expansion of $ln(\mathbf X)$ or better, of $ln(\mathbf I + (\mathbf X
 - \mathbf I))$ when this is convergent.

*Concerning the matrices commutation
You shall be aware that the product of two Matrices $\mathbf A$ and
$\mathbf B$, in general does not commute. Same for the product of 
$ln(\mathbf A)$ and $\mathbf B$.
You can "define" $\mathbf A
   ^{\mathbf B}$ as you like, as far as you state the properties you
expect therefrom.
