Question

Suppose \[AB = \left[ {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{4}}\\{ - {\bf{2}}}&{\bf{3}}\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{3}}\\{\bf{2}}&{\bf{1}}\end{array}} \right]\]. Find A.

Short answer

The matrix \[A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{13}\\{ - 8}&{27}\end{array}} \right]\].

Step-by-step solution

Use the inverse formula

\[{\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]when \[ad - bc \ne 0\].

Find the inverse of B

\[\begin{array}{c}{B^{ - 1}} = {\left[ {\begin{array}{*{20}{c}}7&3\\2&1\end{array}} \right]^{ - 1}}\\ = \frac{1}{{7\left( 1 \right) - 3\left( 2 \right)}}\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\ = \frac{1}{{7 - 6}}\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\ = 1\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\end{array}\]

Find A

Right multiply \[{B^{ - 1}}\] on both sides of \[AB = \left[ {\begin{array}{*{20}{c}}7&3\\2&1\end{array}} \right]\]. Therefore,

\[\begin{array}{c}AB{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}5&4\\{ - 2}&3\end{array}} \right]{B^{ - 1}}\\A = \left[ {\begin{array}{*{20}{c}}5&4\\{ - 2}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&7\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5 - 8}&{ - 15 + 28}\\{ - 2 - 6}&{6 + 21}\end{array}} \right]\\A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{13}\\{ - 8}&{27}\end{array}} \right]\end{array}\]