$$$\left[\begin{array}{cc}2 - \lambda & 1\\1 & 1 - \lambda\end{array}\right]$$$ 的行列式

计算 $$$\left|\begin{array}{cc}2 - \lambda & 1\\1 & 1 - \lambda\end{array}\right|$$$。

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

$$$\left|\begin{array}{cc}2 - \lambda & 1\\1 & 1 - \lambda\end{array}\right| = \left(2 - \lambda\right)\cdot \left(1 - \lambda\right) - \left(1\right)\cdot \left(1\right) = \lambda^{2} - 3 \lambda + 1$$$

$$$\left|\begin{array}{cc}2 - \lambda & 1\\1 & 1 - \lambda\end{array}\right| = \lambda^{2} - 3 \lambda + 1$$$A