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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.

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  • $\begingroup$ Did you really mean $\left( \begin{array}{ccccc} d_0 & -e_0 & 0 &0 & 0 \\ 0 & d_1 & -e_1 & 0 & 0 \\ 0 &0 & d_2 & -e_2 & 0 \\ 0 & 0 & 0 & d_3 & -e_3 \end{array} \right) \left( \begin{array}{c} t_1 \\ t_2 \\ t_3 \\ t_4 \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$ ? The way you wrote it doesn't make sense, because the left side of the equation is a matrix, and the right side is a vector. $\endgroup$ – Nick May 13 '19 at 23:17
  • $\begingroup$ Quite possibly! My original system of equations is $ \begin{matrix} t_0*d_0 - t_1*e_0 = 0 \\ t_1*d_1 - t_2*e_1 = 0 \\ t_2*d_2 - t_3*e_2 = 0 \\ t_3*d_3 - t_4*e_3 = 0 \\ \end{matrix} $ Isn't what you wrote equivalent to this? And isn't the matrix in the post the same thing as well, but aligned so that all t_n reside in their respective columns? $\endgroup$ – Gabe Krause May 13 '19 at 23:40
  • $\begingroup$ Assuming the form I'm after is this, $$ \left[ \begin{matrix} d_0 & -e_0 & 0 & 0 & 0 \\ 0 & d_1 & -e_1 & 0 & 0 \\ 0 & 0 & d_2 & -e_2 & 0 \\ 0 & 0 & 0 & d_3 & -e_3 \\ \end{matrix} \right] \left[ \begin{matrix} t_0 \\ 20 \\ t_2 \\ t_3 \\ t_4 \\ \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right] $$ How do I deal with that constant $ t_1 = 20 $ when solving $ \vec t = A^{-1} * B $ in something like Octave or Matlab? $\endgroup$ – Gabe Krause May 13 '19 at 23:58
  • $\begingroup$ And here's one of the obvious things I could be presenting incorrectly. Are the following two representations drastically different in meaning? $$ \left[ \mathbf A \right] \left[ \mathbf t \right] = \left[ \mathbf B \right] $$ and $$ \left( \mathbf A \right) \left( \mathbf t \right) = \left( \mathbf B \right) $$ $\endgroup$ – Gabe Krause May 14 '19 at 0:19
  • $\begingroup$ No. They are exactly the same. It is simply a matter of personal taste whether people like to use square brackets or round brackets for matrices and vectors. $\endgroup$ – Nick May 14 '19 at 0:20
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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 $t_4$.

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  • $\begingroup$ I like that! And appreciate your attention on this, Nick. I would definitely back-substitute by hand on a system this small. I probably should have mentioned that this was a tiny representation of a larger system, like $$t_n, \; where \; n \gt 1000$$ I am hoping to understand this in a way that would allow me to programmatically plug a large set of $ d_n $ and $ e_n $ values in to something like GNU Octave. But first-things-first. I hope to understand if this can fit into the $Ax=B$ format so that I can use Octave's standard x = linsolve (A, b) function. $\endgroup$ – Gabe Krause May 14 '19 at 0:53
  • $\begingroup$ Considering how I presented the question, this is a fine answer by @Nick . Now that I am more clear about what I need to ask on a forum for this, I've created a more concise question here $\endgroup$ – Gabe Krause May 14 '19 at 18:13

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