$$$\left[\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right]$$$의 행렬식

$$$\left|\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right|$$$을(를) 계산하세요.

2x2 행렬의 행렬식은 $$$\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = a d - b c$$$입니다.

$$$\left|\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right| = \left(\frac{4}{5} - \lambda\right)\cdot \left(\frac{7}{10} - \lambda\right) - \left(\frac{3}{10}\right)\cdot \left(\frac{1}{5}\right) = \lambda^{2} - \frac{3 \lambda}{2} + \frac{1}{2}$$$

$$$\left|\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right| = \frac{\left(\lambda - 1\right) \left(2 \lambda - 1\right)}{2} = 0.5 \left(\lambda - 1\right) \left(2 \lambda - 1\right)$$$A