## Applying Kirchoof's Laws Using Matrices

We can analyse circuits with several voltages and resistors, using Kirchoffs' Laws.
Th Voltage Law: The voltage around a loop in the circuit add up to zero.
The Current Law: The sum of currents into any junction in a circuit adds up to zero.
Consider the circuit below.

Apply the current Law to junction 1.
$I_3-I_1-I_2=0 \rightarrow I_1+I_2-I_3=0$

Applying the voltage law to the left hand loop, and
$V=IR$
to each resistor gives
$5-4_I-3I_3=0 \rightarrow 4I_1+3I_3=5$
.
Applying the voltage law to the right hand loop, and
$V=IR$
to each resistor gives
$3I_3+5I_2-6=0 \rightarrow 5I_2+3I_3=6$
.
We have three simultaneous equations.
$I_1+I_2-I_3=0$

$4I_1+3I_3=5$

$5I_2+3I_3=6$

We can write this in matrix form as
$\left( \begin{array}{ccc} 1 & 1 & -1 \\ 4 & 0 & 3 \\ 0 & 5 & 3 \end{array} \right) \begin{pmatrix}I_1 \\ I_2 \\ I_3 \end{pmatrix}= \begin{pmatrix}0\\5\\6\end{pmatrix}$

Hence
$\begin{pmatrix}I_1\\I_2\\I_3\end{pmatrix}={\left( \begin{array}{ccc} 1 & 1 & -1 \\ 4 & 0 & 3 \\ 0 & 5 & 3 \end{array} \right)}^{-1} \begin{pmatrix}0\\5\\6\end{pmatrix}= \begin{pmatrix} {22/47} \\ {27/47} \\ {49/47} \end{pmatrix}$