Answer
$F_s=\frac{107}{35} y$
Work Step by Step
We have data points $(0,0),(3,-10),(5,-14),(6,-19)$
The matrices can be formed as:
$A=\begin{bmatrix}
0\\
3\\
5\\
6
\end{bmatrix}$
$x=\begin{bmatrix}
a
\end{bmatrix}$
$b=\begin{bmatrix}
0\\
-10\\
14\\
-19
\end{bmatrix}$
Apply matrices to the least square solution:
$x_0=(A^TA)^{-1}A^Tb\\
=(\begin{bmatrix}
0 & 3 & 5 & 6 \end {bmatrix} \begin{bmatrix} 0\\
3 \\
5\\
6 \end {bmatrix})^{-1}\begin{bmatrix}
0 & 3 & 5 & 6 \end {bmatrix}\begin{bmatrix} 0\\
-10 \\
14\\
-19 \end {bmatrix} \\
=\frac{1}{70} \begin{bmatrix} 0 & 3 & 5 & 6 \end{bmatrix} \begin{bmatrix} 0\\
-10 \\
14 \\
-19 \end {bmatrix} \\
=\frac{1}{70} (-214) \\
=\frac{-107}{35} $
According to Hooke's law we obtain:
$F_s=-ky $
We have $a=\frac{-107}{35} \rightarrow k=-a=\frac{107}{35} $
Hence, $F_s=\frac{107}{35} y$
