If I was given a matrix $\mathbf{A}$ and said to obtain the matrix exponential
$$e^{\mathbf{A}t}=\Phi(t)\Phi(0)^{-1}$$
does this matrix $\mathbf{A}$ has only one matrix exponential? Or does it depend on the solution vectors obtained?
I am asking this, because I was solving a problem and the exponential matrix I obtained was different than the one in the instructor's solutions manual. I don't know if this question has been asked here before, I tried looking it up but didn't find anything.
For example, I was asked to obtain the matrix exponential for
$$\mathbf{A}=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$
the answer I obtained was
$$e^{t}\begin{bmatrix} 1 & 2t & 4t^{2} +5t \\ 0 & 1 & 4t \\ 0 & 0 & 1 \end{bmatrix}$$
however, the one in the solutions manual was
$$e^{t}\begin{bmatrix} 1 & 2t & 4t^{2} +3t \\ 0 & 1 & 4t \\ 0 & 0 & 1 \end{bmatrix}$$
Best Answer
$\exp tA$ is a well-defined function of $tA$. It only depends on $tA$.
The Jordan Decomposition is $$ \begin{bmatrix} 1&2&3\\0&1&4\\0&0&1\end{bmatrix} =\begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix} \begin{bmatrix} 1&1&0\\0&1&1\\0&0&1\end{bmatrix} \begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix}^{-1} $$ Next $$ \begin{bmatrix} 1&1&0\\0&1&1\\0&0&1\end{bmatrix}^n =\begin{bmatrix} 1&\binom{n}{1}&\binom{n}{2}\\0&1&\binom{n}{1}\\0&0&1\end{bmatrix} $$ Therefore, $$ \exp t\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1\end{bmatrix} =e^t\begin{bmatrix} 1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix} $$ Thus, $$ \begin{align} \exp t\begin{bmatrix} 1&2&3\\0&1&4\\0&0&1\end{bmatrix} &=e^t\begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix} \begin{bmatrix} 1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix} \begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix}^{-1}\\[6pt] &=e^t\begin{bmatrix} 1&2t&3t+4t^2\\0&1&4t\\0&0&1\end{bmatrix} \end{align} $$ So my answer agrees with the book.
Without seeing your work, I cannot comment on where the problem might be.