The matrix \(\mathbf { A }\), given by
$$\mathbf { A } = \left( \begin{array} { r r r r }
1 & - 1 & 0 & 2 \\
3 & - 1 & 4 & 0 \\
5 & - 8 & - 6 & 19 \\
- 2 & 3 & 2 & - 7
\end{array} \right) ,$$
represents a transformation from \(\mathbb { R } ^ { 4 }\) to \(\mathbb { R } ^ { 4 }\).
- Find the rank of \(\mathbf { A }\) and show that \(\left\{ \left( \begin{array} { r } 2 \\ 2 \\ - 1 \\ 0 \end{array} \right) , \left( \begin{array} { l } 1 \\ 3 \\ 0 \\ 1 \end{array} \right) \right\}\) is a basis for the null space of the transformation.
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- Show that if
$$\mathbf { A x } = p \left( \begin{array} { r }
1 \\
3 \\
5 \\
- 2
\end{array} \right) + q \left( \begin{array} { r }
- 1 \\
- 1 \\
- 8 \\
3
\end{array} \right) ,$$
where \(p\) and \(q\) are given real numbers, then
$$\mathbf { x } = \left( \begin{array} { c }
p + 2 \lambda + \mu \\
q + 2 \lambda + 3 \mu \\
- \lambda \\
\mu
\end{array} \right) ,$$
where \(\lambda\) and \(\mu\) are real numbers.
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- Find the values of \(p\) and \(q\) such that
$$p \left( \begin{array} { r }
1 \\
3 \\
5 \\
- 2
\end{array} \right) + q \left( \begin{array} { r }
- 1 \\
- 1 \\
- 8 \\
3
\end{array} \right) = \left( \begin{array} { r }
3 \\
7 \\
18 \\
- 7
\end{array} \right)$$
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- Find the solution of the equation \(\mathbf { A x } = \left( \begin{array} { r } 3 \\ 7 \\ 18 \\ - 7 \end{array} \right)\) of the form \(\mathbf { x } = \left( \begin{array} { l } 4 \\ 9 \\ \alpha \\ \beta \end{array} \right)\), where \(\alpha\) and \(\beta\) are positive integers to be found.
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