행렬 조건 연립방정식

$A ^{2} - A + E = O$ $( A + E ) ( A - 2 E ) + 3 E = O$ $( A + E ) ( A - 2 E ) = - 3 E$ $\displaystyle ( A + E ) \left\{ - \frac{1}{3} ( A - 2 E ) \right\} = E$이므로 $\displaystyle ( A + E ) ^{- 1} = - \frac{1}{3} ( A - 2 E )$이다. $\displaystyle ( A + E ) {\begin{pmatrix} x \\ y \end{pmatrix}} = {\begin{pmatrix} 3 \\ 6 \end{pmatrix}}$에서 $\displaystyle {\begin{pmatrix} x \\ y \end{pmatrix}} = ( A + E ) ^{- 1} {\begin{pmatrix} 3 \\ 6 \end{pmatrix}}$$\displaystyle = - \frac{1}{3} ( A - 2 E ) {\begin{pmatrix} 3 \\ 6 \end{pmatrix}}$ $\displaystyle = ( - A + 2 E ) {\begin{pmatrix} 1 \\ 2 \end{pmatrix}}$ $\displaystyle = - A {\begin{pmatrix} 1 \\ 2 \end{pmatrix}} + 2 E {\begin{pmatrix} 1 \\ 2 \end{pmatrix}}$ $\displaystyle = {\begin{pmatrix} - 3 \\ 1 \end{pmatrix}} + {\begin{pmatrix} 2 \\ 4 \end{pmatrix}} = {\begin{pmatrix} - 1 \\ 5 \end{pmatrix}}$ $\alpha = - 1 , \beta = 5$ $\therefore \alpha + \beta = 4$