# Is this a linear system of equations?

1. May 19, 2013

### nde

Hello everyone!

I have a question on whether a system of equations can be classified as linear. I have the following matrix:

$\left[ \begin{array}{c} S_t(1) \\ S_t(2) \\ \vdots \\ S_t(\omega_N) \end{array} \right] = \begin{bmatrix} f(x_1, x_2, 1) & f(x_2, x_3, 1) & \cdots & f(x_i, x_{i+1}, 1) \\ f(x_1, x_2, 2) & f(x_2, x_3, 2) & \cdots & f(x_i, x_{i+1}, 2) \\ \vdots & \vdots & \ddots & \vdots \\ f(x_1, x_2, \omega_N) & f(x_2, x_3, \omega_N) & \cdots & f(x_i, x_{i+1}, \omega_N) \\ \end{bmatrix} \times \left[ \begin{array}{c} S_1 \\ S_2 \\ \vdots \\ S_i \end{array} \right] \label{equationsystem}$

where $f(x_i, x_{i+1}, \omega_N)$ is a non-linear function containing two exponential terms and $S_i$ is unknown. Does this system of equations qualify as linear if I know $x_i, x_{i+1}$ and $\omega_N$ and plug it into $f(x_i, x_{i+1}, \omega_N)$ to yield a numerical value (real number)?

If this is true, I should be able to figure out $S_i$ by taking the inverse of the function marix and multiplying both sides with it.

Kind regards.

Last edited: May 19, 2013
2. May 19, 2013

### Staff: Mentor

Yes if the f(...) are considered constant coefficients then you can do the matrix inversion but if they are somehow dependent on the S unknowns then all bets are off.

3. May 19, 2013

### nde

Thanks for your reply. What do you mean when you say that the $S_i$ are somehow dependent on the $f$? Could you please illustrate it with a simple example?

4. May 19, 2013