역행렬로 푸는 연립방정식

$\displaystyle {\begin{pmatrix} a + 1 & & a \\ 1 & & 1 \end{pmatrix}} {\begin{pmatrix} x \\ y \end{pmatrix}} = {\begin{pmatrix} - 4 \\ 1 \end{pmatrix}}$에서 $\displaystyle {\begin{pmatrix} x \\ y \end{pmatrix}} = {\begin{pmatrix} a + 1 & & a \\ 1 & & 1 \end{pmatrix}} ^{- 1} {\begin{pmatrix} - 4 \\ 1 \end{pmatrix}}$$\displaystyle = {\begin{pmatrix} 1 & & - a \\ - 1 & & a + 1 \end{pmatrix}} {\begin{pmatrix} - 4 \\ 1 \end{pmatrix}}$$\displaystyle = {\begin{pmatrix} - 4 - a \\ 4 + a + 1 \end{pmatrix}}$ $\displaystyle = {\begin{pmatrix} - a - 4 \\ a + 5 \end{pmatrix}}$ $x = - a - 4$, $y = a + 5$가 $x + 2 y - 4 a = 0$을 만족시키므로 $( - a - 4 ) + 2 ( a + 5 ) - 4 a = 0$, $- 3 a + 6 = 0$ $\therefore a = 2$