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If \(A = \left[ {\begin{array}{*{20}{c}} 1&5\\ 6&2 \end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}} 3&7\\ 8&4 \end{array}} \right].A{B^T}\) is equal to
1. \(\left[ {\begin{array}{*{20}{c}} {38}&{28}\\ {32}&{56} \end{array}} \right]\)
2. \(\left[ {\begin{array}{*{20}{c}} 3&{40}\\ {42}&8 \end{array}} \right]\)
3. \(\left[ {\begin{array}{*{20}{c}} {43}&{27}\\ {34}&{50} \end{array}} \right]\)
4. \(\left[ {\begin{array}{*{20}{c}} {38}&{32}\\ {28}&{56} \end{array}} \right]\)

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Correct Answer - Option 1 : \(\left[ {\begin{array}{*{20}{c}} {38}&{28}\\ {32}&{56} \end{array}} \right]\)
\(\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 1&5\\ 6&2 \end{array}} \right]\;\left[ B \right] = \left[ {\begin{array}{*{20}{c}} 3&7\\ 8&4 \end{array}} \right]\\ A{B^T} = \left[ {\begin{array}{*{20}{c}} 1&5\\ 6&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&8\\ 7&4 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 + 35}&{8 + 20}\\ {18 + 14}&{48 + 8} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {38}&{28}\\ {32}&{56} \end{array}} \right] \end{array}\)

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