FP1 June 2010 Q6
6. Write down the \(2 \times 2\) matrix that represents
Hence, or otherwise,
Given that \(\mathbf{AB}\) represents the same transformation as \(\mathbf{T}\),
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathbf{T} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ 0 & -8 \end{pmatrix}\) | M1 A1 |
| (2) |
Notes
(c) M1: Accept multiplication of their matrices either way round (this can be implied by correct answer)
A1: cao
| Scheme | Marks |
|---|---|
| \(\mathbf{AB} = \begin{pmatrix} 6 & 1 \\ 4 & 2 \end{pmatrix}\begin{pmatrix} k & 1 \\ c & -6 \end{pmatrix} = \begin{pmatrix} 6k + c & 0 \\ 4k + 2c & -8 \end{pmatrix}\) | M1 A1 A1 |
| (3) |
Notes
(d) M1: Correct matrix multiplication method implied by one or two correct terms in correct positions.
A1: for three correct terms in correct positions
2nd A1: for all four terms correct and simplified
| Scheme | Marks |
|---|---|
| “\(6k + c = 8\)” and “\(4k + 2c = 0\)” Form equations and solve simultaneously | M1 |
| \(k = 2\) and \(c = -4\) | A1 |
| (2) | |
| 9 marks |
Alternative method for (e)
M1: \(\mathbf{AB} = \mathbf{T} \Rightarrow \mathbf{B} = \mathbf{A}^{-1}\mathbf{T} =\) and compare elements to find \(k\) and \(c\). Then A1 as before.Notes
(e) M1: follows their previous work but must give two equations from which \(k\) and \(c\) can be found and there must be attempt at solution getting to \(k =\) or \(c =\).
A1: is cao (but not cso - may follow error in position of \(4k + 2c\) earlier).