Computing characteristic polynomial of $4 \times 4$ matrix

Question

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 $(\lambda - 1)^4$.

Answer 1

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

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.

Answer 2

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