7 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 $$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & 4 \\ 2 & 1 & 4 & 11 \\ 3 & 4 & 1 & 9 \\ 4 & - 3 & 18 & 37 \end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 1 & 1 & 1 & - 1 \\ 2 & 3 & 0 & 1 \\ 3 & 4 & 1 & 0 \\ 4 & 5 & 2 & 0 \end{array} \right)$$ respectively. The null space of \(\mathrm { T } _ { 1 }\) is denoted by \(K _ { 1 }\) and the null space of \(\mathrm { T } _ { 2 }\) is denoted by \(K _ { 2 }\). Show that the dimension of \(K _ { 1 }\) is 2 and that the dimension of \(K _ { 2 }\) is 1 . Find the basis of \(K _ { 1 }\) which has the form \(\left\{ \left( \begin{array} { c } p \\ q \\ 1 \\ 0 \end{array} \right) , \left( \begin{array} { l } r \\ s \\ 0 \\ 1 \end{array} \right) \right\}\) and show that \(K _ { 2 }\) is a subspace of \(K _ { 1 }\).

# Question 7:

## Part: Reduces M₁ to echelon form
$\begin{pmatrix}1&1&1&4\\2&1&4&11\\3&4&1&9\\4&-3&18&37\end{pmatrix}\rightarrow\cdots\rightarrow\begin{pmatrix}1&1&1&4\\0&-1&2&3\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 |

## Part: Finds Dim(K₁)
$\text{Dim}(K_1) = 4 - 2 = 2$ | A1 | (AG)

## Part: Reduces M₂ to echelon form
$\begin{pmatrix}1&1&1&-1\\2&3&0&1\\3&4&1&0\\4&5&2&0\end{pmatrix}\rightarrow\cdots\rightarrow\begin{pmatrix}1&1&1&-1\\0&1&-2&3\\0&0&0&1\\0&0&0&0\end{pmatrix}$ | A1 | (aef)

## Part: Finds Dim(K₂)
$\text{Dim}(K_2) = 4 - 3 = 1$ | A1 | (AG)

## Part: Obtains basis for K₁
$x + y + z + 4t = 0$
$-y + 2z + 3t = 0$
Legitimately obtains basis for $K_1$ is $\left\{\begin{pmatrix}-3\\2\\1\\0\end{pmatrix},\begin{pmatrix}-7\\3\\0\\1\end{pmatrix}\right\}$ | M1, A1A1 | (OE)

## Part: Obtains basis for K₂
$x + y + z - t = 0$
$y - 2z + 3t = 0$
$t = 0$
Legitimately obtains basis for $K_2$ is $\left\{\begin{pmatrix}-3\\2\\1\\0\end{pmatrix}\right\}$ | M1, A1 | (OE) $\Rightarrow K_2 \subset K_1$

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7 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

$$\mathbf { M } _ { 1 } = \left( \begin{array} { r r r r } 
1 & 1 & 1 & 4 \\
2 & 1 & 4 & 11 \\
3 & 4 & 1 & 9 \\
4 & - 3 & 18 & 37
\end{array} \right) \quad \text { and } \quad \mathbf { M } _ { 2 } = \left( \begin{array} { r r r r } 
1 & 1 & 1 & - 1 \\
2 & 3 & 0 & 1 \\
3 & 4 & 1 & 0 \\
4 & 5 & 2 & 0
\end{array} \right)$$

respectively. The null space of $\mathrm { T } _ { 1 }$ is denoted by $K _ { 1 }$ and the null space of $\mathrm { T } _ { 2 }$ is denoted by $K _ { 2 }$. Show that the dimension of $K _ { 1 }$ is 2 and that the dimension of $K _ { 2 }$ is 1 .

Find the basis of $K _ { 1 }$ which has the form $\left\{ \left( \begin{array} { c } p \\ q \\ 1 \\ 0 \end{array} \right) , \left( \begin{array} { l } r \\ s \\ 0 \\ 1 \end{array} \right) \right\}$ and show that $K _ { 2 }$ is a subspace of $K _ { 1 }$.

\hfill \mbox{\textit{CAIE FP1 2012 Q7 [10]}}