normal equations of $ y(t) = \gamma e^{\lambda t} $ for minimizing the error Let $ y(t) = \gamma e^{\lambda t} $ and we have the points $(0,2)\ (1,0.7)\ (3, 0.3)$. The task is to get the parameter so that error is minimal.
So we need to get the matrix for the normal equations. Am i right that in this case is this:
$$
A=\begin{pmatrix}
1 & 1\\
1 & e\\
1 & e^2
\end{pmatrix}
$$
Rest will be $A^TAx = A^Ty$ to solve. But i am not sure about A. Is it correct like this?

Before starting with the LS machinery, please note that the data $(t,y)$ are nonlinearly related (through the exponential); to apply LS is then necessary  to linearize the  data introducing the regression
$$\ln y= \ln (\gamma e^{\lambda t})= \ln \gamma + \lambda t $$  
on the set of data $(t,\ln y)=\{(0, \ln2), (1, \ln0.7), (3, \ln 0.3)\}$.
The LS estimators for the parameters $\ln \gamma$ and $\lambda$ are explicitly given here, and the exercise is done.
