Easy expression, I’m stuck here

Question

This is an expression obtained for resolving a circuit. I tried for 1 hour now and I’m close but I’m doing something wrong.

I’m trying to find detA. (details under the photo)

I’m new to this so be kind with me :)

detA where A is the first matrice

$((1/x)+(1/y))*((1/y)+(1/z))$

I came up with this: $(1/R1*R2)+(1/R1*R3)+(1/R2^2)+(1/R2*R3)-(-1/R2)^2 =$ $(1/R1)((1/R2)+(1/R3)) + (1/R2*R3)$

(note, maybe the parentheses around fractions are useless) There are some similarities, it can be random.

Answer 1

Recall that

$$AV=J \implies V=A^{-1}J$$

and

$$A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}\implies A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d & -b \\ -c & a \end{pmatrix}$$

with

$$\det (A)=\left(\frac1{R_1}+\frac1{R_2}\right)\left(\frac1{R_2}+\frac1{R_3}\right)-\frac1{R_2^2}=\frac1{R_1R_2}+\frac1{R_2R_3}+\frac1{R_3R_1}=\frac{R_1+R_2+R_3}{R_1R_2R_3}$$

Answer 2

Your arrow is incorrect. The equation shows $Ax=b$ where $x$ is the vector of $V(1)$ and $V(2)$, and $b$ is the vector of $J_1$ and $J_2$. The next line shows $x=A^{-1} b$.

To get the inverse of a 2x2 matrix, negate the off diagonals, switch the diagonals, and divide by the determinant. Presumably, the matrix they show would be equivalent to this.