Answer: If \(\alpha=9\), the rank is \(2\).

If \(\alpha\neq9\), the rank is \(3\).

For \(\alpha=9\), a basis for the null space is

\(\left\{\begin{pmatrix}5\\3\\1\\0\end{pmatrix},\begin{pmatrix}1\\4\\0\\-1\end{pmatrix}\right\}.\)

Row reduction gives an echelon form equivalent to

\(\begin{pmatrix}1&-1&-2&-3\\0&-1&3&-4\\0&0&0&\alpha-9\\0&0&0&0\end{pmatrix}.\)

If \(\alpha\neq9\), there are three non-zero rows, so the rank is \(3\).

If \(\alpha=9\), the third row becomes zero, so there are two non-zero rows and the rank is \(2\).

Now take \(\alpha=9\) and solve the homogeneous system. The echelon equations are

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

\(-y+3z-4t=0.\)

From the second equation,

\(y=3z-4t.\)

Substitute into the first equation:

\(x=(3z-4t)+2z+3t=5z-t.\)

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

\(\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=s\begin{pmatrix}5\\3\\1\\0\end{pmatrix}+u\begin{pmatrix}-1\\-4\\0\\1\end{pmatrix}.\)

Equivalently, using a scalar multiple for the second basis vector,

\(\left\{\begin{pmatrix}5\\3\\1\\0\end{pmatrix},\begin{pmatrix}1\\4\\0\\-1\end{pmatrix}\right\}\)

is a basis for the null space.