Answer: (i) \(\operatorname{rank}(\mathbf A)=2\), and a basis for the null space is

\(\left\{\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix},\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}\right\}.\)

(ii) \(\mathbf A\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix}=\begin{pmatrix}3\\21\\24\\27\end{pmatrix}\), so

\(\mathbf x=\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix}+\lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix}+\mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix},\quad \lambda,\mu\in\mathbb R.\)

(i) Row-reducing \(\mathbf A\) gives

\(\begin{pmatrix}1&-1&-2&3\\5&-3&-4&25\\6&-4&-6&28\\7&-5&-8&31\end{pmatrix}\sim\begin{pmatrix}1&-1&-2&3\\0&1&3&5\\0&0&0&0\\0&0&0&0\end{pmatrix}.\)

There are two non-zero rows, so \(\operatorname{rank}(\mathbf A)=2\).

For the null space, write \(\mathbf x=(x,y,z,t)^T\). The reduced equations are

\(x-y-2z+3t=0,\qquad y+3z+5t=0.\)

Let \(z=\lambda\) and \(t=\mu\). Then \(y=-3\lambda-5\mu\), and substituting into the first equation gives \(x=-\lambda-8\mu\). Hence

\(\mathbf x=\lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix}+\mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}.\)

So a basis for the null space is

\(\left\{\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix},\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix}\right\}.\)

(ii) Direct multiplication gives

\(\mathbf A\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix}=\begin{pmatrix}3\\21\\24\\27\end{pmatrix}.\)

Therefore \(\begin{pmatrix}-1&1&-1&1\end{pmatrix}^T\) is one particular solution. Add the whole null space to this particular solution:

\(\mathbf x=\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix}+\lambda\begin{pmatrix}-1\\-3\\1\\0\end{pmatrix}+\mu\begin{pmatrix}-8\\-5\\0\\1\end{pmatrix},\quad \lambda,\mu\in\mathbb R.\)