전치행렬의 성질 판정

행렬 $\displaystyle A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} , B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$라 하면 Ⅰ. $\displaystyle ( A ^{t} ) ^{t} = {\begin{pmatrix} a & c \\ b & d \end{pmatrix}} ^{t} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} = A$ (참) Ⅱ. $\displaystyle ( A + A ^{t} ) ^{t} = \left\{ \begin{array}{lll} \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a & c \\ b & d \end{pmatrix} \end{array} \right\} ^{t}$$\displaystyle = \begin{pmatrix} 2 a & b + c \\ b + c & 2 d \end{pmatrix}$$= A + A ^{t}$ (참) Ⅲ. $\displaystyle ( AB ) ^{t} = \left\{ \begin{array}{ll} \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix} \end{array} \right\} ^{t}$$\displaystyle = \begin{pmatrix} ae + bg & ce + dg \\ af + bh & cf + dh \end{pmatrix}$ $\displaystyle A ^{t} B ^{t} = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} e & g \\ f & h \end{pmatrix}$$\displaystyle = \begin{pmatrix} ae + cf & ag + ch \\ be + df & bg + dh \end{pmatrix}$ (거짓)