How have these two simultaneous equations been solved?

Question

I have two equations where...

$-A_1\omega L+ A_2R=0$

and

$A_1R+A_2\omega L=V_M$

The book I'm using has managed to get them into the forms:

$A_1=\frac{RV_M}{R^2+\omega ^2L^2}$

$A_2=\frac{\omega LV_M}{R^2+\omega ^2L^2}$.

So far I've managed to get $A_1$ as the subject of both but this doesn't seem to help

$A_1=\frac{A_2R}{\omega L}$

$A_1=\frac{V_M-A_2\omega L}{R}$.

Am I on the right tracks? If I am does anyone know what I can do next? If not could someone tell me how to start again?

Thanks

Answer 1

Assuming $R, \omega, L, V_M$ are constants, let $X = \omega L$. You need to solve $$ \begin{pmatrix} -X & R \\ R & X \end{pmatrix} \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} = \begin{pmatrix} 0 \\ V_M\end{pmatrix} $$ If $R = 0$, the solution is trivial ($A_1=0, A_2=V_M/X$).

Assume $R \ne 0$. Multiply the bottom equation by $X/R$ and add to the top, replacing the bottom equation: $$ \begin{pmatrix} -X & R & 0\\ R & X & V_M\end{pmatrix} \to \begin{pmatrix} -X & R & 0\\ 0 & R+X^2/R & XV_M/R\end{pmatrix}, $$ which implies $$ A_2 = \frac{XV_M/R}{R + X^2/R} = \frac{XV_M}{R^2+X^2} $$

Can you finish this?