역행렬 등식 보기 판단

ㄱ. $\left\{ ( A + B ) A ^{-1} \right\} ( A - B ) = ( E + BA ^{-1} ) ( A - B )$ $= A + B - B - BA ^{-1} B = A - BA ^{-1} B$ $\left\{ ( A - B ) A ^{-1} \right\} ( A + B ) = ( E - BA ^{-1} ) ( A + B )$ $= A - B + B - BA ^{-1} B = A - BA ^{-1} B$ ∴ $( A + B ) A ^{-1} ( A - B ) = ( A - B ) A ^{-1} ( A + B )$ (참) ㄴ. $AB ^{2} = ( AB ) B = E$이면 $( AB ) ^{-1} = B$ ∴ $B ^{-1} A ^{-1} = B$ (참) ㄷ. 반례) $\displaystyle A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $\displaystyle B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$이면 $B ^{2} = E$이므로 $AB ^{2} = B ^{2} A$이지만 $AB ^{} \ne B ^{} A$이다. (거짓)