Answer: (i) The rank of the matrix is \(2\). A basis for the range space can be taken as two independent columns of \(\mathbf{M}\), for example \(\left\{\begin{pmatrix}2\\2\\6\\10\end{pmatrix},\begin{pmatrix}-1\\0\\-2\\-3\end{pmatrix}\right\}\).

(ii) A basis for the null space is \(\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix},\begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}\).

We row-reduce \(\mathbf{M}\):

\(\begin{pmatrix}2 & -1 & 1 & 3\\ 2 & 0 & 0 & 5\\ 6 & -2 & 2 & 11\\ 10 & -3 & 3 & 19\end{pmatrix}\)

Subtract the first row from the second, and use the first row to eliminate below:

\(\begin{pmatrix}2 & -1 & 1 & 3\\ 0 & 1 & -1 & 2\\ 0 & 1 & -1 & 2\\ 0 & 2 & -2 & 4\end{pmatrix}\)

Now subtract the second row from the third, and twice the second row from the fourth:

\(\begin{pmatrix}2 & -1 & 1 & 3\\ 0 & 1 & -1 & 2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}\)

There are 2 non-zero rows, so the rank is \(2\).

For the range space, a basis is given by any 2 independent columns of the original matrix. Taking the first two columns:

\(\left\{\begin{pmatrix}2\\2\\6\\10\end{pmatrix},\begin{pmatrix}-1\\0\\-2\\-3\end{pmatrix}\right\}\)

These are independent, so they form a basis for the range space.

For the null space, let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\\t\end{pmatrix}\) and solve \(\mathbf{M}\mathbf{x}=\mathbf{0}\). From the reduced matrix we get

\(2x-y+z+3t=0\)

\(y-z+2t=0\)

From the second equation, \(y=z-2t\). Substitute into the first:

\(2x-(z-2t)+z+3t=0\)

so

\(2x+5t=0\), hence \(x=-\frac{5}{2}t\).

Let \(z=s\) and \(t=u\). Then

\(\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=\begin{pmatrix}-\frac{5}{2}u\\s-2u\\s\\u\end{pmatrix}=s\begin{pmatrix}0\\1\\1\\0\end{pmatrix}+u\begin{pmatrix}-\frac{5}{2}\\-2\\0\\1\end{pmatrix}.\)

Multiplying the second vector by 2 gives an equivalent basis with integer entries:

\(\left\{\begin{pmatrix}0\\1\\1\\0\end{pmatrix},\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix}\right\}\)

So a basis for the null space is \(\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix},\begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}\).