0
$\begingroup$

I have been tasked with solving a series of linear equations in the form of Ax = b, where b is the zero vector. Specifically, I have a matrix E which is 4x7 and a column vector r which is 7x1, as pictured:The E matrix

enter image description here

I must solve for Er = 0.I'm having difficulty finding a function in GNU Octave that will solve this equation for me. I went to the documentation page at https://octave.org/doc/v4.0.0/index.html#SEC_Contents and failed to find any functions that would accept column vectors with variable elements.

I found nothing of use in 1.2.4-Solving Systems of Linear Equations.

linsolve() looked promising, but it doesn't work with variables as far as I can tell.

I found nothing of use under 16-matrix manipulation, 18-linear algebra, nor 25-optimization.

Any advice or suggestions would be appreciated.

$\endgroup$
1
$\begingroup$

You don't need a computer program to solve this. You can almost do it in your head.

The last equation is satisfied by any values of $x, y, z$ so you can forget about it.

If you multiply out the first row of the matrix product, you get $$1 + 2.4 + x + 0 -5.6 + 0 + 0 = 0$$ which is pretty easy to solve for $x$.

The second row is just as easy to solve for $y$, and the third row gives you the value of $z$.

If you have several sets of equations like this to solve with different coefficients in the $E$ matrix, you can rearrange the equations in the form $$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} c_{11} & c_{12} & c_{13} & c_{14} \\ c_{21} & c_{22} & c_{23} & c_{24} \\ c_{31} & c_{32} & c_{33} & c_{34} \end{bmatrix} \begin{bmatrix}d_1\\d_2\\d_3\\d_4\end{bmatrix}$$ and you should then be able to find some routines to multiply the $C$ and $D$ matrices and then solve the equations.


A different way to solve this by computer is to consider all 7 elements of the vector as variables, and then add more equations which give them their known values. That would give a set of 7 equations like $$\begin{bmatrix} 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ 2.3 & 3 & 0 & 1 & 2 & 0 & 3 \\ 0.8 & 0.5 & 2 & 1 & 1 & 2 & 0 \\ 0.3 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ x \\ y \\ c \\ -z \\ d\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 2.4 \\ -5.6 \end{bmatrix}$$ where the three additional equations fix the valued of $a, b, c$ (and the fourth equation, which was irrelevant for finding $x, y, z$, will give the value of $d$).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.