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 \(q\).

## Question 8:

**Reduces matrices to echelon form:**

| $\begin{pmatrix}1 & -2 & 3 & 5\\3 & -4 & 17 & 33\\5 & -9 & 20 & 36\\4 & -7 & 16 & 29\end{pmatrix} \rightarrow \begin{pmatrix}1 & -2 & 3 & 5\\0 & 1 & 4 & 9\\0 & 0 & 1 & 2\\0 & 0 & 0 & 0\end{pmatrix}$ | M1, A1 | Row reduction to echelon form |

| Second matrix similarly reduced to echelon form with final row of zeros | A1 | |

**Finds basis for each null space:**

| $x - 2y + 3z + 5t = 0$, $y + 4z + 9t = 0$, $z + 2t = 0 \Rightarrow \begin{pmatrix}1\\1\\2\\-1\end{pmatrix}$ | M1A1 | |

| $x - 2y - 3t = 0$, $y + 2t = 0$, $z + t = 0 \Rightarrow \begin{pmatrix}1\\2\\1\\-1\end{pmatrix}$ | A1 | 6 marks total |

**Writes $\mathbf{x}_1$ and $\mathbf{x}_2$ appropriately:**

| $\mathbf{x}_1 = \begin{pmatrix}1\\2\\3\\4\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\2\\-1\end{pmatrix}$ and $\mathbf{x}_2 = \begin{pmatrix}1\\2\\3\\4\end{pmatrix} + \mu\begin{pmatrix}1\\2\\1\\-1\end{pmatrix}$ | B1$\checkmark$ | |

**Finds difference:**

| $\mathbf{x}_1 - \mathbf{x}_2 = \begin{pmatrix}\lambda-\mu\\\lambda-2\mu\\2\lambda-\mu\\-\lambda+\mu\end{pmatrix} \Rightarrow \lambda - 2\mu = 5$ and $2\lambda - \mu = 7$ | M1 | |

| $\Rightarrow \lambda = 3$ and $\mu = -1$ | A1 | |

**And solves:**

| $x_1 - x_2 = (4\ 5\ 7\ -4)^T \Rightarrow p = 4$ and $q = -4$ | A1 | 4 marks; **[10 total]** |

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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 [10]}}