Finding Characteristic Polynomial of 4x4 Matrix I need help finding the characteristic polynomial for this symmetric $4\times 4$ matrix:
$$
A=        \begin{pmatrix}
        1275 & -169 & 0 & -208 \\
        -169 & 1531 & -208 & -208 \\
        0 & -208 & 1275 & -256 \\
        -208 & -208 & -256 & 1444\\
        \end{pmatrix}
$$
 My professor says there is a number of linear combinations/row operations that can make finding the characteristic polynomial for this matrix much, much easier than the otherwise tedious way of finding it by expanding the determinant. I've tried everything but I can't find a way to do it...can someone help me?? Thanks so much in advance!
 A: Permuting rows two and three, as well as columns two and three yields the matrix
$$
A_2:=
        \begin{pmatrix}
        1275 & 0 & -169 & -208 \\
        0 &  1275 &-208 & -256 \\
        -169  & -208 & 1531& -208 \\
        -208  & -256& -208 & 1444\\
        \end{pmatrix}.
$$
Denoting $B:=\pmatrix{-169  & -208\\-208  & -256}$, the matrix can be written in block form
$$
A_2 = \pmatrix{ 1275\ I_2 & B \\ B & B + 1700\ I_2} = 1275\ I_4 + \pmatrix{ 0 & B \\ B & B + 425 I_2}.
$$
The matrix $B=\pmatrix{-13\cdot 13  & -13\cdot 16\\-13\cdot 16  & -16\cdot 16}$,
has eigenvalues $0$ and $-425=-256-169$ with eigenvectors vectors $v_0=\pmatrix{16\\-13}$ and $v_{-425}=\pmatrix{13\\16}$, respectively.
These vectors combined yield the eigenvectors of $A_2$ (pairs eigenvector $\to$ eigenvalue)
$$
\pmatrix{v_0\\0} \to 1275, \ 
\pmatrix{0\\v_0} \to 1700, \ 
\pmatrix{v_{-425}\\v_{-425}} \to 850, \ 
\pmatrix{v_{-425}\\-v_{-425}} \to 1700.
$$
Thus, the characteristic polynomial is
$$
p(\lambda) = (\lambda-1700)^2 (\lambda-1275) (\lambda-850).
$$
Expanding the characteristic polynomial the traditional way is no fun, the determinant of the matrix is $\approx 3\cdot 10^{12}$.
