Finding determinant of $3\times3$ matrix $$A = \left(\begin{matrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{matrix}\right)$$
$$\det A = \begin{vmatrix}A\end{vmatrix} = 
(\lambda - 1) \begin{vmatrix}
\lambda - 3 & 1 \\
1 & \lambda + 1 \\
\end{vmatrix} + 1\begin{vmatrix}
1 & 1 \\
-3 & \lambda + 1 \\
\end{vmatrix} -
1\begin{vmatrix}
1 & \lambda - 3 \\
-3 & 1 \\
\end{vmatrix}\\
= ((\lambda - 1)(\lambda - 3)(\lambda + 1)) - (1 + (\lambda + 1) + 3) + (1 +3\lambda - 9) \\
= (\lambda - 1)(\lambda - 3)(\lambda + 1) + (\lambda + 1) + 2(\lambda - 3)$$
The solution says that it is $(\lambda - 2)(\lambda + 2)(\lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution. 

$$=(\lambda - 1)\Big((\lambda - 3)(\lambda + 1)-1\Big) + ((\lambda + 1) + 3) -(1 +3\lambda - 9)  $$
$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) + (\lambda + 4) -(3\lambda - 8)  $$
$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) - 2\lambda +12  $$
$$=\lambda ^3 - 3\lambda^2 -2\lambda +4 - 2\lambda +12  $$
$$=\lambda ^3 - 3\lambda^2 -4\lambda +16$$
