대칭 행렬 성분

[출제의도] 행렬의 정의를 이해하여 문제해결하기 $a _{ij} + a _{ji} = 0$에서 $a _{ij} = - a _{ji}$이므로 $a _{11} = 0$, $a _{12} = - a _{21}$, $a _{22} = 0$ $\displaystyle A = {\begin{pmatrix} a _{11} & a _{12} \\ a _{21} & a _{22} \end{pmatrix}} = {\begin{pmatrix} 0 & a _{12} \\ - a _{12} & 0 \end{pmatrix}}$ $b _{ij} - b _{ji} = 0$에서 $b _{ij} = b _{ji}$이므로 $b _{12} = b _{21}$ $\displaystyle B = {\begin{pmatrix} b _{11} & b _{12} \\ b _{21} & b _{22} \end{pmatrix}} = {\begin{pmatrix} b _{11} & b _{12} \\ b _{12} & b _{22} \end{pmatrix}}$ $\displaystyle 2 A - B = 2 {\begin{pmatrix} 0 & a _{12} \\ - a _{12} & 0 \end{pmatrix}} - {\begin{pmatrix} b _{11} & b _{12} \\ b _{12} & b _{22} \end{pmatrix}}$ $\displaystyle = {\begin{pmatrix} - b _{11} & 2 a _{12} - b _{12} \\ - 2 a _{12} - b _{12} & - b _{22} \end{pmatrix}} = \left( \begin{array}{ll} {\begin{matrix} 1 & 2 \\ - 2 & 4 \end{matrix}} \end{array} \right)$ $- b _{11} = 1$에서 $b _{11} = - 1$, $- b _{22} = 4$에서 $b _{22} = - 4$ $2 a _{12} - b _{12} = 2$ $\cdots \cdots \text{㉠}$, $- 2 a _{12} - b _{12} = - 2$ $\cdots \cdots \text{㉡}$ ㉠, ㉡에서 $a _{12} = 1$, $b _{12} = 0$ $\therefore$ $\displaystyle A = {\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}}$, $\displaystyle B = {\begin{pmatrix} - 1 & 0 \\ 0 & - 4 \end{pmatrix}}$ $\displaystyle A ^{2} - B = {\begin{pmatrix} - 1 & 0 \\ 0 & - 1 \end{pmatrix}} - {\begin{pmatrix} - 1 & 0 \\ 0 & - 4 \end{pmatrix}} = {\begin{pmatrix} 0 & 0 \\ 0 & 3 \end{pmatrix}}$ 따라서 행렬 $A ^{2} - B$의 $\left( 2 , 2 \right)$성분은 $3$