Correct Answer - Option 4 :
\(\frac{{\left( {s + 4} \right)}}{{\left( {{s^2} - s - 4} \right)}}\)
From the given state space equations,
\(A = \left[ {\begin{array}{*{20}{c}}
1&2\\
2&0
\end{array}} \right]\)
\(B = \left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right]\)
\(C = \left[ {\begin{array}{*{20}{c}}
1&0
\end{array}} \right]\)
Transfer function = C[sI - A]-1 B + D
\(sI - A = \left[ {\begin{array}{*{20}{c}}
s&0\\
0&s
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
1&2\\
2&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{s - 1}&{ - 2}\\
{ - 2}&s
\end{array}} \right]\)
\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{{s^2} - s - 4}}\left[ {\begin{array}{*{20}{c}}
s&2\\
2&{s - 1}
\end{array}} \right]\)
Transfer function \(= \left[ {\begin{array}{*{20}{c}}
1&0
\end{array}} \right]\frac{1}{{\left( {{s^2} - s - 4} \right)}}\left[ {\begin{array}{*{20}{c}}
s&2\\
2&{s - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right]\)
\(= \frac{1}{{\left( {{s^2} - s - 4} \right)}}\left[ {\begin{array}{*{20}{c}}
s&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right]\)
\(= \frac{{s + 4}}{{\left( {{s^2} - s - 4} \right)}}\)