I need how to get to the answer for this question. I made this in a study group, but we didn't write down how we got to the answer and now i've been stuck on it for a long time. i cant seem to make any sense of it.
we have 8 unknown coefficients to determine. We shall see that the $x$-coefficients can be determined independently from $y$-coefficients, and that the procedure is exactly the same for both sets of coefficients. We will therefore focus on the $x$-coefficients. The equations that $x(t)$ has to satisfy are: $$ \left\{\begin{aligned} x(0) & =x_1, \\ x^{\prime}(0) & =3\left(x_2-x_1\right), \\ x^{\prime}(1) & =3\left(x_4-x_3\right), \\ x(1) & =x_4 . \end{aligned}\right. $$ Write down the matrix $A$ and the right-hand side vector $b$ correspoinding to the system of equations (19). Solve the system (19) using Gaussian elimination to find the explicit formulas for $\left(p_x, q_x, r_x, s_x\right) \in$ $\mathbb{R}^4$ in terms of the data $\left(x_1, x_2, x_3, x_4\right) \in \mathbb{R}^4$. Clearly the formulas for $\left(p_y, q_y, r_y, s_y\right) \in \mathbb{R}^4$ are analogous. $$
This is the answer we got:
$$ \begin{gathered} x(0)=p_x=x_1 \\ x^{\prime}(0)=q_x=3\left(x_2-x_1\right) \\ x^{\prime}(1)=q_x+2 r_x+3 s_x=3\left(x_4-x_3\right) \\ x(1)=p_x+q_x+r_x+s_x=x_4 \end{gathered} $$ This is the part that makes me confused, how does x'(0) correspond $=q_x=3\left(x_2-x_1\right) $ and so on.
$$ \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \begin{gathered} x_1 \\ 3\left(x_2-x_1\right) \\ -6 x_2+3 x_1+3 x_3 \\ x_4-3 x_3+3 x_2-x_1 \end{gathered} $$