FP3 June 2012 Q8
8. The matrix \(\mathbf{M}\) is given by \[\mathbf{M} = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ -1 & 0 & 4 \end{pmatrix}\]
The straight line \(l_1\) is mapped onto the straight line \(l_2\) by the transformation represented by the matrix \(\mathbf{M}\).
The equation of \(l_1\) is \((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = 0\), where \(\mathbf{a} = 3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) and \(\mathbf{b} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}\).
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix} 2-\lambda & 1 & 0 \\ 1 & 2-\lambda & 0 \\ -1 & 0 & 4-\lambda \end{vmatrix} = 0 \therefore (2-\lambda)(2-\lambda)(4-\lambda) - (4-\lambda) = 0\) | M1 |
| \((4-\lambda) = 0\) verifies \(\lambda = 4\) is an eigenvalue (can be seen anywhere) | M1 |
| \(\therefore (4-\lambda)\left\{4 - 4\lambda + \lambda^2 - 1\right\} = 0 \quad \therefore (4-\lambda)\left\{\lambda^2 - 4\lambda + 3\right\} = 0\) | A1 |
| \(\therefore (4-\lambda)(\lambda - 1)(\lambda - 3) = 0\) and 3 and 1 are the other two eigenvalues | M1 A1 |
| (5) |
Notes
a1M1: Condone missing = 0. (They might expand the determinant using any row or column)
a2M1: Shows \(\lambda = 4\) is an eigenvalue. Some working needed need to see = 0 at some stage.
a1A1: Three term quadratic factor cao, may be implicit (this A depends on 1st M only)
a2M1: Attempt at factorisation (usual rules), solving to \(\lambda =\) .
a2A1: cao. If they state \(\lambda = 1\) and 3 please give the marks.
| Scheme | Marks |
|---|---|
| Set \(\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ -1 & 0 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = 4\begin{pmatrix} x \\ y \\ z \end{pmatrix}\) or \(\begin{pmatrix} -2 & 1 & 0 \\ 1 & -2 & 0 \\ -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) | M1 |
| Solve \(-2x + y = 0\) and \(x - 2y = 0\) and \(-x = 0\) to obtain \(x = 0,\ y = 0,\ z = k\) | M1 |
| Obtain eigenvector as \(\mathbf{k}\) (or multiple) | A1 |
| (3) |
Notes
b1M1: Using \(\mathrm{A}\boldsymbol{x} = 4\boldsymbol{x}\) o.e.
b2M1: Getting a pair of correct equations.
b1A1: cao
| Scheme | Marks |
|---|---|
| \(l_1\) has equation which may be written \(\begin{pmatrix} 3 + \lambda \\ 2 - \lambda \\ -2 + 2\lambda \end{pmatrix}\) | B1 |
| So \(l_2\) is given by \(\mathbf{r} = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ -1 & 0 & 4 \end{pmatrix}\begin{pmatrix} 3 + \lambda \\ 2 - \lambda \\ -2 + 2\lambda \end{pmatrix}\) | M1 |
| i.e. \(\mathbf{r} = \begin{pmatrix} 8 + \lambda \\ 7 - \lambda \\ -11 + 7\lambda \end{pmatrix}\) | M1 A1 |
| So \((\mathbf{r} - \mathbf{c}) \times \mathbf{d} = \mathbf{0}\) where \(\mathbf{c} = 8\mathbf{i} + 7\mathbf{j} - 11\mathbf{k}\) and \(\mathbf{d} = \mathbf{i} - \mathbf{j} + 7\mathbf{k}\) | A1ft |
| (5) | |
| (13 marks) |
Notes
c1B1: Using \(\mathbf{a}\) and \(\mathbf{b}\).
c1M1: Using \(\mathbf{r} = \mathbf{M} \times\) their matrix in \(\mathbf{a}\) and \(\mathbf{b}\).
c2M1: Getting an expression for \(l_2\) with at least one component correct.
c1A1: cao all three components correct
c2A1ft: ft their vector, must have \(\mathbf{r} =\) or \((\mathbf{r} - \mathbf{c}) \times \mathbf{d} = 0\) need both equation and \(\mathbf{r}\).