Show that $0$ has multiplicity $3$ in $M-4I$? Question from an exam:
Consider the matrix $M=$
\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}

Show that $4$ is an eigen value of the above matrix with multiplicity $3$.

I considered $M-4I$ from which I got 
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
By elementary row operations:
$$M-4I=$$
\begin{bmatrix} 1&1&1&1&1&1\\0&0&0&0&0&0\\0&0&0&0&0&0\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
So $0$ is an eigen value with multiplicity at least $2$ since $M-2I$ has two zero rows.

How to show that $0$ has multiplicity $3$ in $M-4I$?

If one can show how to proceed after this,I will be really grateful.
Please help 

Since $M$ is symmetric, the geometric and algebraic multiplicity of its eigenvalues coincide. 
If you finish your row reduction, you get 
$$
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
\xrightarrow{\ \ \ \ \ \ \ \ \ \ }
\begin{bmatrix} 1&1&1&0&0&-1\\0&0&0&1&0&1\\0&0&0&0&1&1\\0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\end{bmatrix}
$$
As there are three leading ones, this shows that the set of solutions of $(M-4I)x=0$ has three dependent solutions, and three independent. Thus $\dim \ker(M-4I)=3$. This means that the eigenvalue 4 has geometric and algebraic multiplicity 3.
