CAIE FP1 2013 June — Question 8

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices

8 The linear transformations \(\mathrm { T } _ { 1 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) and \(\mathrm { T } _ { 2 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) are represented by the matrices \(\mathbf { M } _ { 1 }\) and \(\mathbf { M } _ { 2 }\) respectively, where $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & - 2 & 3 & 5 \\ 3 & - 4 & 17 & 33 \\ 5 & - 9 & 20 & 36 \\ 4 & - 7 & 16 & 29 \end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 1 & - 2 & 0 & - 3 \\ 2 & - 1 & 0 & 0 \\ 4 & - 7 & 1 & - 9 \\ 6 & - 10 & 0 & - 14 \end{array} \right) .$$ The null spaces of \(\mathrm { T } _ { 1 }\) and \(\mathrm { T } _ { 2 }\) are denoted by \(K _ { 1 }\) and \(K _ { 2 }\) respectively. Find a basis for \(K _ { 1 }\) and a basis for \(K _ { 2 }\). It is given that \(\mathbf { a } = \left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)\). The vectors \(\mathbf { x } _ { 1 }\) and \(\mathbf { x } _ { 2 }\) are such that \(\mathbf { M } _ { 1 } \mathbf { x } _ { 1 } = \mathbf { M } _ { 1 } \mathbf { a }\) and \(\mathbf { M } _ { 2 } \mathbf { x } _ { 2 } = \mathbf { M } _ { 2 } \mathbf { a }\). Given that \(\mathbf { x } _ { 1 } - \mathbf { x } _ { 2 } = \left( \begin{array} { c } p \\ 5 \\ 7 \\ q \end{array} \right)\), find \(p\) and \(q\).

8 The linear transformations $\mathrm { T } _ { 1 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ and $\mathrm { T } _ { 2 } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ are represented by the matrices $\mathbf { M } _ { 1 }$ and $\mathbf { M } _ { 2 }$ respectively, where

$$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 
1 & - 2 & 3 & 5 \\
3 & - 4 & 17 & 33 \\
5 & - 9 & 20 & 36 \\
4 & - 7 & 16 & 29
\end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 
1 & - 2 & 0 & - 3 \\
2 & - 1 & 0 & 0 \\
4 & - 7 & 1 & - 9 \\
6 & - 10 & 0 & - 14
\end{array} \right) .$$

The null spaces of $\mathrm { T } _ { 1 }$ and $\mathrm { T } _ { 2 }$ are denoted by $K _ { 1 }$ and $K _ { 2 }$ respectively. Find a basis for $K _ { 1 }$ and a basis for $K _ { 2 }$.

It is given that $\mathbf { a } = \left( \begin{array} { l } 1 \\ 2 \\ 3 \\ 4 \end{array} \right)$. The vectors $\mathbf { x } _ { 1 }$ and $\mathbf { x } _ { 2 }$ are such that $\mathbf { M } _ { 1 } \mathbf { x } _ { 1 } = \mathbf { M } _ { 1 } \mathbf { a }$ and $\mathbf { M } _ { 2 } \mathbf { x } _ { 2 } = \mathbf { M } _ { 2 } \mathbf { a }$. Given that $\mathbf { x } _ { 1 } - \mathbf { x } _ { 2 } = \left( \begin{array} { c } p \\ 5 \\ 7 \\ q \end{array} \right)$, find $p$ and $q$.

\hfill \mbox{\textit{CAIE FP1 2013 Q8}}