Determinant of $$$\left[\begin{array}{ccc}1 - \lambda & 1 & -1\\-3 & - \lambda - 3 & 3\\-2 & -2 & 2 - \lambda\end{array}\right]$$$

Calculate $$$\left|\begin{array}{ccc}1 - \lambda & 1 & -1\\-3 & - \lambda - 3 & 3\\-2 & -2 & 2 - \lambda\end{array}\right|$$$.

Subtract column $$$2$$$ multiplied by $$$1 - \lambda$$$ from column $$$1$$$: $$$C_{1} = C_{1} - \left(1 - \lambda\right) C_{2}$$$.

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

Add column $$$2$$$ to column $$$3$$$: $$$C_{3} = C_{3} + C_{2}$$$.

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

Expand along row $$$1$$$:

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

The determinant of a 2x2 matrix is $$$\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = a d - b c$$$.

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

Finally, $$$\left(-1\right)\cdot \left(\lambda^{3}\right) = - \lambda^{3}$$$.

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