If I have the matrix down below how can I get rid of I1 so I can have everything in terms of I2?
$ \begin{pmatrix} (10+s) & (-s-6) \\ (-s-6) & (s+\dfrac{4}{s}+6) \\ \end{pmatrix} $ $ \begin{pmatrix} I1\\ I2\\ \end{pmatrix} = \begin{pmatrix} \dfrac{6}{s+3}\\ \dfrac{-6}{s}-1\\ \end{pmatrix} $
This is Laplace Transform so I can have everything in terms of $s$.
Reducing by row, observing that writing $a=s+6$, $b=\frac{6}{s+3}$ and $c=-\frac{6}{s}-1$ the augmented matrix becomes $$ \left( \begin{array}{cc|c} 4+a & -a & b\\ -a & \frac{4}{s}+a & c \end{array} \right)\xrightarrow[]{R_2\leftrightarrow R_2+R_1} \left( \begin{array}{cc|c} 4+a & -a & b\\ 4 & \frac{4}{s} & b+c \end{array} \right)\\ \xrightarrow[]{R_1\leftrightarrow \frac{R_1}{4+a}} \left( \begin{array}{cc|c} 1 & -\frac{a}{4+a} & \frac{b}{4+a}\\ 4 & \frac{4}{s} & b+c \end{array} \right)\xrightarrow[]{R_2\leftrightarrow R_2-4R_1}\left( \begin{array}{cc|c} 1 & -\frac{a}{4+a} & \frac{b}{4+a}\\ 0 & \frac{4}{s} + \frac{4a}{4+a} & b+c-\frac{4b}{4+a} \end{array} \right) $$ So you have
\begin{align*} I_2&=\frac{b+c-\frac{4b}{4+a}}{\frac{4}{s} + \frac{4a}{4+a} }\\ I_1&=\frac{b}{4+a}+\frac{a}{4+a}I_2 \end{align*}