Answer: (i) The rank of the matrix is \(2\).

(ii) A basis for the null space is \(\left\{\begin{pmatrix}-2\\0\\1\\1\end{pmatrix},\begin{pmatrix}-3\\1\\0\\0\end{pmatrix}\right\}\).

Let the matrix of \(T\) be \(M\).

(i) Rank of \(M\)

Row-reducing \(M\) gives an echelon form with two non-zero rows, for example

\(\begin{pmatrix}1 & 3 & -2 & 4\\ 5 & 15 & -9 & 19\\ -2 & -6 & 3 & -7\\ 3 & 9 & -5 & 11\end{pmatrix} \sim \begin{pmatrix}1 & 3 & -2 & 4\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}.\)

There are two pivot rows, so the rank is \(2\).

(ii) Basis for the null space

To find the null space, solve \(M\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=\mathbf{0}\). The echelon form corresponds to the equations

\(x+3y-2z+4t=0,\qquad z-t=0.\)

From \(z-t=0\), we have \(z=t\). Substitute into the first equation:

\(x+3y-2t+4t=0 \implies x=-3y-2t.\)

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

\(\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=\begin{pmatrix}-3s-2u\\s\\u\\u\end{pmatrix}=s\begin{pmatrix}-3\\1\\0\\0\end{pmatrix}+u\begin{pmatrix}-2\\0\\1\\1\end{pmatrix}.\)

So a basis for the null space is \(\left\{\begin{pmatrix}-3\\1\\0\\0\end{pmatrix},\begin{pmatrix}-2\\0\\1\\1\end{pmatrix}\right\}\).