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.