Concept Used:Application:We have,\(B=\left( {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right) =\left( {\begin{array}{*{20}{c}} 9&-4\\ -2&1 \end{array}} \right)\)and,\(A=\left( {\begin{array}{*{20}{c}} w&x\\ y&z \end{array}} \right) =\left( {\begin{array}{*{20}{c}} 1&4\\ 2&5 \end{array}} \right) \)Now,(aw + by) = (9 × 1) + (-4 × 2) = 1(ax + bz) = (9 × 4) + (-4 × 5) = 16(cw + dy) = (-2 × 1) + (1 × 2) = 0(cx + dz) = (-2 × 4) + (1 × 5) = -3Hence, BA = \(\left( {\begin{array}{*{20}{c}} 1&16\\ 0&-3 \end{array}} \right)\)