Computing characteristic polynomial of $4 times 4$ matrix

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: $$D=begin{vmatrix} lambda-1 &3&0 \ 2&lambda+6&lambda-7 \ 1&4&lambda-4 end{vmatrix}=begin{vmatrix} lambda-1 &3&0 \ 0&lambda-2 &1-lambda\ 1&0&lambda-4 end{vmatrix}.$$

$$lambda I- A= begin{bmatrix} lambda -1&3&0&-3 \ 2&lambda + 6&0&-13 \ 0&3&lambda -1&-3 \ 1&4&0&lambda-8 end{bmatrix} $$

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.