Computing characteristic polynomial of $4 \times 4$ matrix Given the following matrix
$$A = \begin{bmatrix} 1 & -3 & 0 & 3 \\ -2 & -6 & 0 & 13 \\ 0 & -3 & 1 & 3 \\ -1 & -4 & 0 & 8\end{bmatrix}$$
I need to calculate $|\lambda I - A|$.
I get to a very complicated determinant. Is there an easier way? The answer says it’s $(\lambda - 1)^4$.
 A: $$\lambda I- A=
\begin{bmatrix}
\lambda -1&3&0&-3 \\
2&\lambda + 6&0&-13 \\
0&3&\lambda -1&-3 \\
1&4&0&\lambda-8
\end{bmatrix}
 $$
By developing the determinant using the third column, we'll have:
$$\det(\lambda I -A)=  (\lambda -1) \cdot \begin{bmatrix}
\lambda -1&3&-3 \\
2&\lambda + 6&-13 \\
1&4&\lambda-8
\end{bmatrix} $$ $$=  (\lambda -1)\left( (\lambda -1)\cdot \begin{bmatrix}\lambda + 6&-13 \\ 4& \lambda-8 \end{bmatrix} - 3 \cdot \begin{bmatrix}2&-13 \\ 1& \lambda-8 \end{bmatrix} 
 +3 \cdot \begin{bmatrix}2&\lambda +6 \\ 1& 4 \end{bmatrix} \right)$$
by solving this, you will get your answer. I don't think it's too complicated to handle. notice that  $$\begin{bmatrix}\lambda + 6&-13 \\ 4& \lambda-8 \end{bmatrix}, \begin{bmatrix}2&-13 \\ 1& \lambda-8 \end{bmatrix} 
, \begin{bmatrix}2&\lambda +6 \\ 1& 4 \end{bmatrix}$$ are three minors which you will have to find their determinant.
A: Hint:
First expand along the 3rd column:
$$\lambda I-A= \begin{vmatrix}
\lambda-1 &3&0&-3 \\ 2&\lambda+6& 0&-13 \\ 0&3&\lambda-1&-3\\ 1&4&0&\lambda-8
\end{vmatrix}=(\lambda-1)\begin{vmatrix}
\lambda-1 &3&-3 \\ 2&\lambda+6& -13 \\   1&4&\lambda-8
\end{vmatrix}$$
Now, either you apply directly Sarrus' rule, or you simplify the 3rd order determinant (we'll denote it $D$) with elementary operations to make a determinant with more zeros:
$$D=\begin{vmatrix}
\lambda-1 &3&0 \\ 2&\lambda+6&\lambda-7  \\ 1&4&\lambda-4
\end{vmatrix}=\begin{vmatrix}
\lambda-1 &3&0 \\ 0&\lambda-2 &1-\lambda\\ 1&0&\lambda-4
\end{vmatrix}.$$
