en forma matricial $$ \begin{equation*} u = \alpha_{0} + \alpha_{1} x = \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} \end{equation*} $$
reemplazando los puntos \( x_{1} \) y \( x_{2} \) $$ \begin{align*} \alpha_{0} + \alpha_{1} x_{1} &= u_{1} \\ \alpha_{0} + \alpha_{1} x_{2} &= u_{2} \end{align*} $$
en forma matricial $$ \begin{equation*} \begin{bmatrix} 1 & x_{1} \\ 1 & x_{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} = \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \end{equation*} $$
resolviendo el sistema $$ \begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} = \begin{bmatrix} \frac{x_{2}}{x_{2} - x_{1}} & -\frac{x_{1}}{x_{2} - x_{1}} \\ -\frac{1}{x_{2} - x_{1}} & \frac{1}{x_{2} - x_{1}} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \end{equation*} $$
reemplazando las incógnitas $$ \begin{align*} u &= \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} \\ &= \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} \frac{x_{2}}{x_{2} - x_{1}} & -\frac{x_{1}}{x_{2} - x_{1}} \\ -\frac{1}{x_{2} - x_{1}} & \frac{1}{x_{2} - x_{1}} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \\ &= \begin{bmatrix} \frac{x_{2}}{x_{2} - x_{1}} - \frac{1}{x_{2} - x_{1}} x & -\frac{x_{1}}{x_{2} - x_{1}} + \frac{1}{x_{2} - x_{1}} x \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \\ &= \begin{bmatrix} N_{1} & N_{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \end{align*} $$
Reescribiendo \( u \) $$ \begin{equation*} u = \bigg( \frac{x_{2}}{x_{2} - x_{1}} - \frac{1}{x_{2} - x_{1}} x \bigg) u_{1} + \bigg( -\frac{x_{1}}{x_{2} - x_{1}} + \frac{1}{x_{2} - x_{1}} x \bigg) u_{2} \end{equation*} $$