linear algebra - How to find characteristic polynomial of this matrix?

Let, $A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}$. Knowing that $4$ is one of its eigenvalues, find the characteristic polynomial of $A$.

Well if $4$ is an eigenvalues of $A$, one should have $|A-4I_{4}|=0$ . And so,

$\begin{vmatrix} 0&0&1&0\\1&-3&1&0\\0&1&-3&0 \\0&0&0&0 \end{vmatrix}=0$

It's clear that the previous equation is true (the determinant of $(A-4I_{4})=0$). Now that the factor $(\lambda-4)$ was pull out, one gets a new matrix by removing the null row and null column.

$A'=\begin{bmatrix} 0&0&1\\1&-3&1\\0&1&-3&\end{bmatrix}$

The characteristic polynomial of $A'$ will be a $3^{th}$ degree polynomial, which product with $(\lambda-4)$ equals to a $4^{th}$ degree polynomial.

Now, in order of finding the characteristic polynomial of $A'$ one must to solve the characteristic equation:

$\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix}=0$

My doubt is on finding this determinant. I already tryed Laplace's transformations in order to make null row or a column, but I couldn't do it.

Can you give me a clue? Thanks.

Answer

...and how to find the charateristic polynomial of the original matrix, $A$

$$\begin{align}\mathrm{det}(A - \lambda \mathrm{I}) &= 0 \tag{1}\\ \begin{vmatrix}(4-\lambda)&0&1&0 \\ 1&(1-\lambda)&1&0\\ 0&1&(1-\lambda)&0\\ 0&0&0&(4-\lambda) \end{vmatrix} &= 0 \tag{2}\\ (4-\lambda)\begin{vmatrix}(4-\lambda)&0&1\\ 1&(1-\lambda)&1\\ 0&1&(1-\lambda) \tag{3}\\ \end{vmatrix} &= 0 \\ (4-\lambda) \left[(4-\lambda) \left[(1-\lambda)^2 -1 \right] + 1\right] &= 0\tag{4} \\ (4-\lambda) \left[(4-\lambda)(\lambda^2 -2\lambda) + 1\right] &= 0\tag{5} \\ (4-\lambda) \left[(4\lambda^2 -8\lambda -\lambda^3+2\lambda^2) + 1\right] &= 0 \tag{6}\\ (4-\lambda)(-\lambda^3 + 6\lambda^2 - 8\lambda +1 ) &= 0 \tag{7}\\ (\lambda -4)(\lambda^3 - 6\lambda^2 + 8\lambda - 1)&= 0\tag{8}\\ \end{align}$$