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 $3,4,4,9,4,2$.
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