$$$\left[\begin{array}{ccc}2 - \lambda & 2 & 2\\2 & 6 - \lambda & 2\\2 & 2 & 2 - \lambda\end{array}\right]$$$의 행렬식

$$$\left|\begin{array}{ccc}2 - \lambda & 2 & 2\\2 & 6 - \lambda & 2\\2 & 2 & 2 - \lambda\end{array}\right|$$$을(를) 계산하세요.

$$$1$$$열에서 $$$2$$$열의 $$$1 - \frac{\lambda}{2}$$$배를 뺍니다: $$$C_{1} = C_{1} - \left(1 - \frac{\lambda}{2}\right) C_{2}$$$

$$$\left|\begin{array}{ccc}2 - \lambda & 2 & 2\\2 & 6 - \lambda & 2\\2 & 2 & 2 - \lambda\end{array}\right| = \left|\begin{array}{ccc}0 & 2 & 2\\- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & 6 - \lambda & 2\\\lambda & 2 & 2 - \lambda\end{array}\right|$$$

열 $$$3$$$에서 열 $$$2$$$를 뺍니다: $$$C_{3} = C_{3} - C_{2}$$$.

$$$\left|\begin{array}{ccc}0 & 2 & 2\\- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & 6 - \lambda & 2\\\lambda & 2 & 2 - \lambda\end{array}\right| = \left|\begin{array}{ccc}0 & 2 & 0\\- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & 6 - \lambda & \lambda - 4\\\lambda & 2 & - \lambda\end{array}\right|$$$

행 $$$1$$$에 대해 전개:

$$$\left|\begin{array}{ccc}0 & 2 & 0\\- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & 6 - \lambda & \lambda - 4\\\lambda & 2 & - \lambda\end{array}\right| = \left(0\right) \left(-1\right)^{1 + 1} \left|\begin{array}{cc}6 - \lambda & \lambda - 4\\2 & - \lambda\end{array}\right| + \left(2\right) \left(-1\right)^{1 + 2} \left|\begin{array}{cc}- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & \lambda - 4\\\lambda & - \lambda\end{array}\right| + \left(0\right) \left(-1\right)^{1 + 3} \left|\begin{array}{cc}- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & 6 - \lambda\\\lambda & 2\end{array}\right| = - 2 \left|\begin{array}{cc}- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & \lambda - 4\\\lambda & - \lambda\end{array}\right|$$$

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

$$$\left|\begin{array}{cc}- \frac{\lambda^{2}}{2} + 4 \lambda - 4 & \lambda - 4\\\lambda & - \lambda\end{array}\right| = \left(- \frac{\lambda^{2}}{2} + 4 \lambda - 4\right)\cdot \left(- \lambda\right) - \left(\lambda - 4\right)\cdot \left(\lambda\right) = \frac{\lambda^{3}}{2} - 5 \lambda^{2} + 8 \lambda$$$

마지막으로, $$$\left(-2\right)\cdot \left(\frac{\lambda^{3}}{2} - 5 \lambda^{2} + 8 \lambda\right) = - \lambda \left(\lambda - 8\right) \left(\lambda - 2\right).$$$

$$$\left|\begin{array}{ccc}2 - \lambda & 2 & 2\\2 & 6 - \lambda & 2\\2 & 2 & 2 - \lambda\end{array}\right| = - \lambda \left(\lambda - 8\right) \left(\lambda - 2\right)$$$A