Answer
$A^{-1}$ does not exist.
Work Step by Step
To find $A^{-1}$, we have $$\left[A\ \ I \right]\left[ \begin {array}{cccccccc} 1&0&3&0&1&0&0&0\\ 0
&2&0&4&0&1&0&0\\ 1&0&3&0&0&0&1&0
\\ 0&2&0&4&0&0&0&1\end {array} \right]
. $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccccc} 1&0&3&0&0&0&1&0\\ 0
&1&0&2&0&0&0&1/2\\ 0&0&0&0&1&0&-1&0
\\ 0&0&0&0&0&1&0&-1\end {array} \right]
. $$
Then we obtain that $A^{-1}$ does not exist.
