OR

The linear transformation \(T:\mathbb R^4\to\mathbb R^4\) is represented by the matrix

\(\mathbf M=\begin{pmatrix}1&-2&3&-4\\2&-4&7&-9\\4&-8&14&-18\\5&-10&17&-22\end{pmatrix}.\)

Find the rank of \(\mathbf M\).

Obtain a basis for the null space \(K\) of \(T\).

Evaluate

\(\mathbf M\begin{pmatrix}1\\-2\\2\\-1\end{pmatrix},\)

and hence show that any solution of

\(\mathbf M\mathbf x=\begin{pmatrix}15\\33\\66\\81\end{pmatrix}\)

has the form

\(\begin{pmatrix}1\\-2\\2\\-1\end{pmatrix}+\lambda\mathbf e_1+\mu\mathbf e_2,\)

where \(\lambda\) and \(\mu\) are scalars and \(\{\mathbf e_1,\mathbf e_2\}\) is a basis for \(K\). Hence obtain a solution \(\mathbf x'\) for which the sum of the components is \(6\) and the sum of the squares of the components is \(26\).