linear algebra - How to find the eigen values of the given matrix

Given the matrix

$$\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3 \end{bmatrix}$$

find its eigen values(preferably by elementary row/column operations).

Since I don't know any other method other than elementary operations to find eigen values so I tried writing the characteristic polynomial of the matrix which is follows:

$$\begin{bmatrix} x-5&-1&-1&-1&-1&-1\\-1&x-5&-1&-1&-1&-1\\-1&-1&x-5&-1&-1&-1\\-1&-1&-1&x-4&-1&0\\-1&-1&-1&-1&x-4&0\\-1&-1&-1&0&0&x-3 \end{bmatrix}$$

Using $R1=R1-(R2+R3+R4+R5+R6)$

$$\begin{bmatrix} x&-x+8&-x+8&-x+6&-x+6&-x+4\\-1&x-5&-1&-1&-1&-1\\-1&-1&x-5&-1&-1&-1\\-1&-1&-1&x-4&-1&0\\-1&-1&-1&-1&x-4&0\\-1&-1&-1&0&0&x-3 \end{bmatrix}$$

Answer

Call your matrix $A$. Let $B=A-3I$.
$$B=\pmatrix{ 2&1&1&1&1&1\\ 1&2&1&1&1&1\\ 1&1&2&1&1&1\\ 1&1&1&1&1&0\\ 1&1&1&1&1&0\\ 1&1&1&0&0&0}.$$
$B$ has two identical columns (4 and 5), so we try to remove one of them. Perform the column operation $C5\leftarrow C5-C4$ followed by the inverse row operation $R4\leftarrow R4+R5$:
$$\pmatrix{ 2&1&1&1&0&1\\ 1&2&1&1&0&1\\ 1&1&2&1&0&1\\ 2&2&2&2&0&0\\ 1&1&1&1&0&0\\ 1&1&1&0&0&0}.$$
We now get a zero eigenvalue at the $(5,5)$-th position. Remove the fifth row and column:
$$\pmatrix{ 2&1&1&1&1\\ 1&2&1&1&1\\ 1&1&2&1&1\\ 2&2&2&2&0\\ 1&1&1&0&0}.$$
The first two rows in this matrix sans $I_2$ are identical, so we try to remove a duplicate row. Do $R1\leftarrow R1-R2$ and then the inverse row operation $C2\leftarrow C2+C1$:
$$\pmatrix{ 1&0&0&0&0\\ 1&3&1&1&1\\ 1&2&2&1&1\\ 2&4&2&2&0\\ 1&2&1&0&0}.$$
We now get another eigenvalue $1$ at the top-left position. Remove the first row and column:
$$\pmatrix{ 3&1&1&1\\ 2&2&1&1\\ 4&2&2&0\\ 2&1&0&0}.$$
Do $R1\leftarrow R1-R2$ and $C2\leftarrow C2+C1$ again:

$$\pmatrix{ 1&0&0&0\\ 2&4&1&1\\ 4&6&2&0\\ 2&3&0&0}.$$
Now we get another eigenvalue $1$ at the top-left position. Remove the first row and column:
$$\pmatrix{ 4&1&1\\ 6&2&0\\ 3&0&0}.$$
We may now calculate its characteristic polynomial by hand. It is
$$\begin{align} &(x-4)(x-2)x - 6x -3(x-2)\\ =\,&x^3 - 6x^2 + 8x - 6x - 3x + 6\\ =\,&x^3 - 6x^2 - x + 6\\ =\,&(x-6)(x^2-1). \end{align}$$
The eigenvalues of this matrix are $6,1,-1$. Therefore the eigenvalues of $B$ are $0,1,1,6,1,-1$ and the eigenvalues of $A=B+3I$ are $3,4,4,9,4,2$.