Answer: \(\operatorname{rank}(M)=2\).
A valid basis for the range space is
\(\left\{\begin{pmatrix}2\\2\\6\\10\end{pmatrix},\begin{pmatrix}-1\\0\\-2\\-3\end{pmatrix}\right\}.\)
A basis for the null space is
\(\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix},\begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}.\)
Row-reduce the matrix:
\(\begin{pmatrix}2&-1&1&3\\2&0&0&5\\6&-2&2&11\\10&-3&3&19\end{pmatrix}\sim\begin{pmatrix}2&-1&1&3\\0&1&-1&2\\0&1&-1&2\\0&2&-2&4\end{pmatrix}\sim\begin{pmatrix}2&-1&1&3\\0&1&-1&2\\0&0&0&0\\0&0&0&0\end{pmatrix}.\)
There are two non-zero rows, so
\(\operatorname{rank}(M)=2.\)
The range space is spanned by the independent columns of \(M\). The first two original columns are independent, so one valid basis is
\(\left\{\begin{pmatrix}2\\2\\6\\10\end{pmatrix},\begin{pmatrix}-1\\0\\-2\\-3\end{pmatrix}\right\}.\)
For the null space, let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\\t\end{pmatrix}\). The reduced equations are
\(2x-y+z+3t=0,\qquad y-z+2t=0.\)
Let \(t=\lambda\) and \(z=\mu\). Then
\(y=\mu-2\lambda,\qquad x=-\frac52\lambda.\)
Thus
\(\mathbf{x}=\lambda\begin{pmatrix}-\frac52\\-2\\0\\1\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\\0\end{pmatrix}.\)
Multiplying the first vector by \(2\), a basis for the null space is
\(\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix},\begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}.\)
