Determinant of $$$\left[\begin{array}{ccc}1 - \lambda & 1 & 3\\1 & 5 - \lambda & 1\\3 & 1 & 1 - \lambda\end{array}\right]$$$

Calculate $$$\left|\begin{array}{ccc}1 - \lambda & 1 & 3\\1 & 5 - \lambda & 1\\3 & 1 & 1 - \lambda\end{array}\right|$$$.

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

$$$\left|\begin{array}{ccc}1 - \lambda & 1 & 3\\1 & 5 - \lambda & 1\\3 & 1 & 1 - \lambda\end{array}\right| = \left|\begin{array}{ccc}0 & 1 & 3\\- \lambda^{2} + 6 \lambda - 4 & 5 - \lambda & 1\\\lambda + 2 & 1 & 1 - \lambda\end{array}\right|$$$

Subtract column $$$2$$$ multiplied by $$$3$$$ from column $$$3$$$: $$$C_{3} = C_{3} - 3 C_{2}$$$.

$$$\left|\begin{array}{ccc}0 & 1 & 3\\- \lambda^{2} + 6 \lambda - 4 & 5 - \lambda & 1\\\lambda + 2 & 1 & 1 - \lambda\end{array}\right| = \left|\begin{array}{ccc}0 & 1 & 0\\- \lambda^{2} + 6 \lambda - 4 & 5 - \lambda & 3 \lambda - 14\\\lambda + 2 & 1 & - \lambda - 2\end{array}\right|$$$

Expand along row $$$1$$$:

$$$\left|\begin{array}{ccc}0 & 1 & 0\\- \lambda^{2} + 6 \lambda - 4 & 5 - \lambda & 3 \lambda - 14\\\lambda + 2 & 1 & - \lambda - 2\end{array}\right| = \left(0\right) \left(-1\right)^{1 + 1} \left|\begin{array}{cc}5 - \lambda & 3 \lambda - 14\\1 & - \lambda - 2\end{array}\right| + \left(1\right) \left(-1\right)^{1 + 2} \left|\begin{array}{cc}- \lambda^{2} + 6 \lambda - 4 & 3 \lambda - 14\\\lambda + 2 & - \lambda - 2\end{array}\right| + \left(0\right) \left(-1\right)^{1 + 3} \left|\begin{array}{cc}- \lambda^{2} + 6 \lambda - 4 & 5 - \lambda\\\lambda + 2 & 1\end{array}\right| = - \left|\begin{array}{cc}- \lambda^{2} + 6 \lambda - 4 & 3 \lambda - 14\\\lambda + 2 & - \lambda - 2\end{array}\right|$$$

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

$$$\left|\begin{array}{cc}- \lambda^{2} + 6 \lambda - 4 & 3 \lambda - 14\\\lambda + 2 & - \lambda - 2\end{array}\right| = \left(- \lambda^{2} + 6 \lambda - 4\right)\cdot \left(- \lambda - 2\right) - \left(3 \lambda - 14\right)\cdot \left(\lambda + 2\right) = \lambda^{3} - 7 \lambda^{2} + 36$$$

Finally, $$$\left(-1\right)\cdot \left(\lambda^{3} - 7 \lambda^{2} + 36\right) = - \lambda^{3} + 7 \lambda^{2} - 36$$$.

$$$\left|\begin{array}{ccc}1 - \lambda & 1 & 3\\1 & 5 - \lambda & 1\\3 & 1 & 1 - \lambda\end{array}\right| = - \lambda^{3} + 7 \lambda^{2} - 36$$$A