6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { A }\), where $$\mathbf { A } = \left( \begin{array} { r r r r } 1 & 2 & - 1 & \alpha \\ 2 & 3 & - 1 & 0 \\ 2 & 1 & 2 & - 2 \\ 0 & 1 & - 3 & - 2 \end{array} \right)$$ Given that the dimension of the range space of T is 4 , show that \(\alpha \neq 1\). It is now given that \(\alpha = 1\). Show that the vectors $$\left( \begin{array} { l } 1 \\ 2 \\ 2 \\ 0 \end{array} \right) , \quad \left( \begin{array} { l } 2 \\ 3 \\ 1 \\ 1 \end{array} \right) \quad \text { and } \quad \left( \begin{array} { r } - 1 \\ - 1 \\ 2 \\ - 3 \end{array} \right)$$ form a basis for the range space of T . Given also that the vector \(\left( \begin{array} { c } p \\ 1 \\ 1 \\ q \end{array} \right)\) is in the range space of T , find a condition satisfied by \(p\) and \(q\).

$\begin{pmatrix} 1 & 2 & -1 & \alpha \\ 0 & -1 & 1 & -2\alpha \\ 0 & -3 & 4 & -2-2\alpha \\ 0 & 1 & -3 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & -1 & \alpha \\ 0 & -1 & 1 & -2\alpha \\ 0 & 0 & 1 & 4\alpha - 2 \\ 0 & 0 & 0 & 6\alpha - 6 \end{pmatrix}$ | M1A1

$\text{Dim} = 4 \Rightarrow \alpha \neq 1$ | A1 | | [3]

$a + 2b - c = 0$
$2a + 3b - c = 0$
$2a + b + 2c = 0$
$b - 3c = 0$ | Show $a = b = c = 0$ | M1 | attempt to solve

$\text{Linearly independent and dim } \mathbb{R}(\mathbb{T}) \text{ not } 4: \text{ basis}$ | A1 | | [2]

$a + 2b - c = p$
$2a + 3b - c = 1$
$2a + b + 2c = 1$
$b - 3c = q$ | Attempt to find $a, b, c$ in terms of $q$ or $p$ | M1A1
| | A1 | | [3]

$6p + q = 3$ | A1

**Alternative solution:**

Use row operations as in (i) | M1

$\text{Final column } \begin{pmatrix} p \\ 1-2p \\ 4p-2 \\ 6p+q-3 \end{pmatrix}$ | A1

$6p + q = 3$ | A1

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6 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { A }$, where

$$\mathbf { A } = \left( \begin{array} { r r r r } 
1 & 2 & - 1 & \alpha \\
2 & 3 & - 1 & 0 \\
2 & 1 & 2 & - 2 \\
0 & 1 & - 3 & - 2
\end{array} \right)$$

Given that the dimension of the range space of T is 4 , show that $\alpha \neq 1$.

It is now given that $\alpha = 1$. Show that the vectors

$$\left( \begin{array} { l } 
1 \\
2 \\
2 \\
0
\end{array} \right) , \quad \left( \begin{array} { l } 
2 \\
3 \\
1 \\
1
\end{array} \right) \quad \text { and } \quad \left( \begin{array} { r } 
- 1 \\
- 1 \\
2 \\
- 3
\end{array} \right)$$

form a basis for the range space of T .

Given also that the vector $\left( \begin{array} { c } p \\ 1 \\ 1 \\ q \end{array} \right)$ is in the range space of T , find a condition satisfied by $p$ and $q$.

\hfill \mbox{\textit{CAIE FP1 2010 Q6 [8]}}