# Solving a linear mass balance equation in the form of Ax = 0 with variables in vector x

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:  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.

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 $$d$$).