Linear Regression: linear or reciprocal function? The problem is given below:

Simultaneous values of time $t$ and output $y$ from a specific sensor has been measured and is tabulated below $$\begin{array}{cc}
t & y \\ 
\hline 
1 & 17 \\ 
2 & 15 \\
3 & 11 \\
4 & 10\\
5&8\\
6&7\\
7&7
\end{array} $$
  Determine whether the model $y_1(t) = \beta_0 + \beta_1 t$ or $y_2(t) = \gamma_0 + \gamma_1t^{-1}$ gives the best least squares fit to the data.

by having the design matrix and observation vector 

In the first model I get following beta: $\beta_0=17.71$ and $\beta_1= -1.75$
by the formula:
$\beta = (X^TX)^{-1}X^Ty$.
My question is then, how to make the design matrix for the second model. the $t^{-1}$ confuse me!
 A: Simply take the second column of your X1 matrix and invert the values, that is it should be $X1(:,2) = [1,1/2,1/3,\ldots,1/7]^T$
Then determine which line gives you a better fit between the two. 
It should be noted, this type of regression is sometimes called "Linear Regression With Basis Functions." This concept can add a tremendous amount of flexibility to linear regression and you can get all kinds of neat fits. If you have interest, a simple search of this will give you many interesting articles/presentations/lecture notes on the topic.
A: Linear function
$$
y_{1}(t) = \beta_{0} + \beta_{1} t
$$
Linear system:
$$
\begin{align}
\mathbf{A} \beta &= y \\
%
\left[
\begin{array}{cc}
 1 & 1 \\
 1 & 2 \\
 1 & 3 \\
 1 & 4 \\
 1 & 5 \\
 1 & 6 \\
 1 & 7 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 \beta_{0} \\
 \beta_{1}
\end{array}
\right]
%
&=
%
\left[ \begin{array}{c}
 17 \\
 15 \\
 11 \\
 10 \\
 8 \\
 7 \\
 7
\end{array}
\right]
%
\end{align}
$$
Solution:
$$
 \beta_{LS} = 
%
\left[
\begin{array}{c}
 \beta_{0} \\
 \beta_{1}
\end{array}
\right]
=
%
\frac{1}{28}
\left[
\begin{array}{r}
  496 \\
  -49
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
 17.7143 \\
 -1.753
\end{array}
\right]
%
$$
Error:
$$
 r = \mathbf{A} \beta - y 
  \qquad \Rightarrow \qquad
 r \cdot r \approx 7.67857
$$


Reciprocal function
$$
y_{2}(t) = \gamma_{0} + \frac{\gamma_{1}} {t}
$$
Linear system:
$$
\begin{align}
\mathbf{A} \beta &= \frac{1}{y} \\
%
\left[
\begin{array}{cc}
 1 & 1 \\
 1 & 2 \\
 1 & 3 \\
 1 & 4 \\
 1 & 5 \\
 1 & 6 \\
 1 & 7 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 \gamma_{0} \\
 \gamma_{1}
\end{array}
\right]
%
&=
%
\left[ \begin{array}{c}
 \frac{1}{17} \\
 \frac{1}{15} \\
 \frac{1}{11} \\
 \frac{1}{10} \\
 \frac{1}{8} \\
 \frac{1}{7} \\
 \frac{1}{7}
\end{array}
\right]
%
\end{align}
$$
Solution:
$$
 \gamma_{LS} = 
%
\left[
\begin{array}{c}
 \gamma_{0} \\
 \gamma_{1}
\end{array}
\right]
=
%
\frac{1}{4398240}
\left[
\begin{array}{c}
 181296 \\
 68891
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
 0.0412201 \\
 0.0156633
\end{array}
\right]
%
$$
Error:
$$
 r = \mathbf{A} \gamma - \frac{1}{y} 
  \qquad \Rightarrow \qquad
 r \cdot r \approx 0.000213205
$$

