For A = \(\left[ {\begin{array}{*{20}{c}} 3&1\\ { - 1}&2 \end{array}} \right]\), then 14A-1 is given by :
1
\(14\left[ {\begin{array}{*{20}{c}} 2&{ - 1}\\ 1&3 \end{array}} \right]\)
2
\(\left[ {\begin{array}{*{20}{c}} 4&{ - 2}\\ 2&6 \end{array}} \right]\)
3
\(2\left[ {\begin{array}{*{20}{c}} 2&{ - 1}\\ 1&{ - 3} \end{array}} \right]\)
4
\(2\left[ {\begin{array}{*{20}{c}} -3&{ - 1}\\ 1&{ - 2} \end{array}} \right]\)