I just was working with matrix exponentials for solving problems in control theory. Suppose $A $ is a square matrix. How can we interpret $A_1 = e^ {\textstyle-A\log(t) }$, where $\log$ is natural logarithm? Is there a formula for extending the scalar case of $e^ {\textstyle-a\log(t) }$ which gives $\dfrac{1}{t^a}$? How do we evaluate $A_1$?
Here is how I proceeded along the lines of scalar case:
$$A_1 = \exp(-A\log t)) = \exp(-\log {t^A}) \overset{\textstyle?}{=}\;( t^A)^{-1}$$
Is the final equality correct?
With the matrix exponential, more care is necessary. One approach is to multiply each item in the matrix $A$ by $\ln t$ and then find the diagonal matrix and take advantage of its structure.
Try working these two examples and see if you get them.
Example 1: Matrix is in diagonal form
$A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$.
For a matrix of this type, we can take advantage of the fact that it is a diagonal matrix, and right away write out:
$\displaystyle e^{-A \ln t} = e^{\begin{bmatrix}-1 & 0\\0 & -1 \end{bmatrix}*\ln t} = \begin{bmatrix} 1/t & 0 \\ 0 & 1/t \end{bmatrix}$
Example 2: Diagonalizable Matrix
$A = \begin{bmatrix}-1 & -3\\-1 & -3\end{bmatrix}$, so
$\displaystyle e^{-A \ln t} = e^{\begin{bmatrix}1 & 3\\1 & 3 \end{bmatrix}*\ln t} = \begin{bmatrix}-3 & 1 \\ 1 & 1 \end{bmatrix} \cdot e^{\begin{bmatrix}0 & 0 \\ 0 & 4 \ln t \end{bmatrix}}\cdot \begin{bmatrix} -1/4 & 1/4 \\ 1/4 & 3/4 \end{bmatrix} = \begin{bmatrix}-3 & 1 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 \\ 0 & t^4 \end{bmatrix}\cdot \begin{bmatrix} -1/4 & 1/4 \\ 1/4 & 3/4 \end{bmatrix} = \begin{bmatrix} \dfrac{t^4}{4}+ \dfrac{3}{4} & \dfrac{3 t^4}{4}-\dfrac{3}{4} \\ \dfrac{t^4}{4}-\dfrac{1}{4} & \dfrac{3 t^4}{4} + \dfrac{1}{4}\end{bmatrix}$
Example 3: Non-diagonalizable Matrix
For this matrix, we will write it using the Jordan Normal Form as $A = P \cdot J \cdot P^{-1}$ and take advantage of the Jordan block.
$A = \begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}$, so
$\displaystyle e^{-A \ln t} = e^{\begin{bmatrix}-1 & -2\\0 &-1 \end{bmatrix}*\ln t} = \begin{bmatrix}1 & 0 \\ 0 & -\dfrac{1}{2} \end{bmatrix} \cdot e^{\begin{bmatrix}-\ln t & \ln t \\ 0 & - \ln t \end{bmatrix}}\cdot \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & -\dfrac{1}{2} \end{bmatrix} \cdot \begin{bmatrix}\dfrac{1}{t} & \dfrac{\ln t}{t} \\ 0 & \dfrac{1}{t} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{t} & -\dfrac{2 \ln t}{t} \\ 0 & \dfrac{1}{t} \end{bmatrix} $
Example 4
$A = \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 2 \\ 2 & 1 & 0\end{bmatrix}$
Give it a go. The matrix is diagonalizable and you should get:
$$e^{-A \ln t} = \frac{1}{15 t^3}\begin{bmatrix}4 t^5+5 t^2+6 & 2 t^5-5 t^2+3 & 6 (1-t^5) \\ ~~2 (2 t^5-5 t^2+3) & 2 t^5+10 t^2+3 & 6 (1-t^5) \\ ~~6 (1-t^5) & 3 (1-t^5) & 3 (3 t^5+2)\end{bmatrix}$$