$$$\left[\begin{array}{cc}1 - \lambda & 1\\1 & 1 - \lambda\end{array}\right]$$$ 的行列式

計算 $$$\left|\begin{array}{cc}1 - \lambda & 1\\1 & 1 - \lambda\end{array}\right|$$$。

2x2 矩陣的行列式為 $$$\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = a d - b c$$$。

$$$\left|\begin{array}{cc}1 - \lambda & 1\\1 & 1 - \lambda\end{array}\right| = \left(1 - \lambda\right)\cdot \left(1 - \lambda\right) - \left(1\right)\cdot \left(1\right) = \lambda^{2} - 2 \lambda$$$

$$$\left|\begin{array}{cc}1 - \lambda & 1\\1 & 1 - \lambda\end{array}\right| = \lambda \left(\lambda - 2\right)$$$A