I'd like to use the $Ax=B$ form for solving the following system. $$ \left[ \begin{matrix} t_0*d_0 & -t_1*e_0 & 0 & 0 & 0 \\ 0 & t_1*d_1 & -t_2*e_1 & 0 & 0 \\ 0 & 0 & t_2*d_2 & -t_3*e_2 & 0 \\ 0 & 0 & 0 & t_3*d_3 & -t_4*e_3 \\ \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right] $$ Where: $ t_n $ are variables. And $ d_{n} $ and $ e_{n} $ are constants.
Clearly this solves if all elements of $ \vec t = 0 $ but I'm looking for another solution, where one element of $ \vec t $ is constrained to a constant, say $t_1 = 20$. Can this be done? Is $Ax=B$ even appropriate here?
Note: I'm revisiting a problem that could benefit from linear algebra after taking a very long hiatus. Apologies for missing obvious things.
Just do simple "back-substitution". The last equation is equivalent to $t_3 = \frac{e_3}{d_3} t_4$. Then the third equation gives $t_2 = \frac{e_2}{d_2} t_3 = \frac{e_2 e_3}{d_2 d_3} t_4$. Keep repeating to write all the variables $t_1,t_2,t_3$ as multiples of $t_4$.