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I have a system of 6 equations with 6 unknowns that can be written in matrix form below: $$ \left[ \begin{array}{cccccc} a_1 & a_2 & a_3 & 0 & 0 & 0\\ 0 & 0 & 0 & a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 & 0 & 0 & 0\\ 0 & 0 & 0 & b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 & 0 & 0 & 0\\ 0 & 0 & 0 & c_1 & c_2 & c_3\\ \end{array}\right].\left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{array}\right]=\left[ \begin{array}{c} r_1 \\ r_2 \\ s_1 \\ s_2 \\ t_1 \\ t_2 \end{array}\right] $$

Where $a,b,c,r,s,t$ are all known.

I can just invert this 6x6 matrix, but that would be a huge pain. Are there any structures/symmetry I can take advantage of here?

I noticed that I can "squash" the matrix like this: $$ \left[ \begin{array}{cccccc} a_1 & a_2 & a_3 & a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 & b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 & c_1 & c_2 & _3\\ \end{array}\right].\left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{array}\right]=\left[ \begin{array}{c} r_1 + r_2 \\ s_1 + s_2 \\ t_1 + t_2 \end{array}\right] $$

This reduces the matrix size by a bit, but as far as I know a non square matrix does not have an inverse, so I'm not sure how to proceed from here?

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  • $\begingroup$ This "reduction" usually changes the solution of your linear equation. $\endgroup$
    – user592521
    Commented Sep 13, 2021 at 8:36
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    $\begingroup$ @GrafZahl Really? How so? From my understanding the reduction should be equivalent? $\endgroup$
    – l3utterfly
    Commented Sep 13, 2021 at 8:37
  • $\begingroup$ It isn't, since you are "throwing away some equations" Think of $\begin{pmatrix}1 &0 \\0&1\end{pmatrix}x=0$. This is not equivalent to $x_1+x_2=0$. $\endgroup$
    – user592521
    Commented Sep 13, 2021 at 8:39

1 Answer 1

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You could define $B=\begin{pmatrix} a_1 &a_2 & a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{pmatrix}$. Then $\begin{pmatrix} x_1\\x_2 \\x_3\end{pmatrix}=B^{-1}\begin{pmatrix} r_1\\ s_1 \\t_1\end{pmatrix}$ and $\begin{pmatrix} x_4\\x_5 \\x_6\end{pmatrix}=B^{-1}\begin{pmatrix} r_2\\ s_2 \\t_2\end{pmatrix}$ .

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