I set up the least squares system to find coefficients $\vec{x} = (x_1, x_2)$ for fitting the model function $$f(t) = x_1t + x_2e^t$$ to the three data points $(1,2), (2,3), (3,5).$ I want to know if the system have a unique solution.
\begin{equation} \begin{split} f(1) = & x_1 t + x_2 e = 2\\ f(2) = & 2x_1 t + x_2 e^2 = 3\\ f(3) = & 3x_1 t + x_2 e^3 = 5 \\ \end{split} \quad \equiv \quad \begin{split} \begin{bmatrix} x_1 + x_2e\\ 2x_1 + x_2e^2 \\ 3x_1 + x_2e^3 \end{bmatrix}= \begin{bmatrix} 2\\3\\5 \end{bmatrix} \end{split} \ \equiv \ \begin{split} \begin{bmatrix} 1 & e\\ 2 & e^2 \\ 3 & e^3 \end{bmatrix} \begin{bmatrix} x_1\\x_2 \end{bmatrix} = \begin{bmatrix} 2\\3\\5 \end{bmatrix} \end{split} \end{equation}
Since minimizing the least squares is equivalent to solving the normal equation, I use it to solve for the solution $$\min\limits_{\vec{x}}\|Ax - b\|_2 \equiv A^\top A \vec{x} = A^\top b = \begin{bmatrix} 1 & 2 & 3\\ e & e^2 & e^3\\ \end{bmatrix} \begin{bmatrix} 1 & e\\ 2 & e^2 \\ 3 & e^3 \end{bmatrix} \vec{x} = \begin{bmatrix} 1 & 2 & 3\\ e & e^2 & e^3\\ \end{bmatrix} \begin{bmatrix} 2\\3\\5 \end{bmatrix} $$ $$ \vec{x} = \begin{bmatrix} 14 & e + 2^e + 3e^3\\ e + 2^e + 3e^3 & e^2 + e^4 + e^6\\ \end{bmatrix}^{-1} \times \begin{bmatrix} 23\\2e + 3e^2 + 5e^3 \end{bmatrix} = \begin{bmatrix} 1.5942\\0.0088 \end{bmatrix}$$
So in order to know if $\vec{x}$ is unique I reduced $A^{m\times n}$ into reduced row echelon form to find the linearly independent rows/columns: RREF$(A) = \text{RREF}\left(\begin{bmatrix} 1 & e\\ 2 & e^2 \\ 3 & e^3 \end{bmatrix} \right) = \begin{bmatrix} 1 & 0\\ 0 & 1 \\ 0 & 0 \end{bmatrix}.$
Rank(A) = $\min(m,n) = 2 \implies A$ is full rank $\implies \vec{x}$ is unique.
I am not sure whether my conclusion about the uniqueness of $\vec{x}$ is true since $m>n$. What I know is that assuming $m\le n$ then my conclusion would have been true but this is the case where $m>n.$ Any suggestions and ideas are welcome.