FP3 June 2011 Q7

EdexcelOld spec12 marksMatrices

7. The matrix \(\mathbf{M}\) is given by \[\mathbf{M} = \begin{pmatrix} k & -1 & 1 \\ 1 & 0 & -1 \\ 3 & -2 & 1 \end{pmatrix}, \qquad k \neq 1\]

(a) Show that \(\det\mathbf{M} = 2 - 2k\). (2)

(b) Find \(\mathbf{M}^{-1}\), in terms of \(k\). (5)

The straight line \(l_1\) is mapped onto the straight line \(l_2\) by the transformation represented by the matrix \(\begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & -1 \\ 3 & -2 & 1 \end{pmatrix}\).

The equation of \(l_2\) is \((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = 0\), where \(\mathbf{a} = 4\mathbf{i} + \mathbf{j} + 7\mathbf{k}\) and \(\mathbf{b} = 4\mathbf{i} + \mathbf{j} + 3\mathbf{k}\).

Scheme	Marks
\(\det\mathbf{M} = k(0 - 2) + 1(1 + 3) + 1(-2 - 0) = -2k + 4 - 2 = 2 - 2k\)	M1 A1
	(2)

Mark scheme (b)

Scheme	Marks
\(\mathbf{M}^{T} = \begin{pmatrix} k & 1 & 3 \\ -1 & 0 & -2 \\ 1 & -1 & 1 \end{pmatrix}\) so cofactors \(= \begin{pmatrix} -2 & -1 & 1 \\ -4 & k - 3 & k + 1 \\ -2 & 2k - 3 & 1 \end{pmatrix}\)	M1
(\(-1\) A mark for each term wrong) \(\mathbf{M}^{-1} = \dfrac{1}{2 - 2k}\begin{pmatrix} -2 & -1 & 1 \\ -4 & k - 3 & k + 1 \\ -2 & 2k - 3 & 1 \end{pmatrix}\)	M1 A3
	(5)

Notes

1M1 Finding matrix of cofactors or its transpose

2M1 Finding inverse matrix, 1/(det) cofactors + transpose

1A1 At least seven terms correct (so at most 2 incorrect) condone missing det

2A1 At least eight terms correct (so at most 1 incorrect) condone missing det

3A1 All nine terms correct, condone missing det

Mark scheme (c)

Scheme	Marks
Let \((x, y, z)\) be on \(l_1\). Equation of \(l_2\) can be written as \(\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 7 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}\).	B1
Use \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}\) with \(k = 2\). i.e. \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \dfrac{1}{-2}\begin{pmatrix} -2 & -1 & 1 \\ -4 & -1 & 3 \\ -2 & 1 & 1 \end{pmatrix}\begin{pmatrix} 4 + 4\lambda \\ 1 + \lambda \\ 7 + 3\lambda \end{pmatrix}\)	M1
\(\therefore \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3\lambda + 1 \\ 4\lambda - 2 \\ 2\lambda \end{pmatrix}\)	M1 A1
and so \((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}\) where \(\mathbf{a} = \mathbf{i} - 2\mathbf{j}\) and \(\mathbf{b} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) or equivalent or \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\) where \(\mathbf{a} = \mathbf{i} - 2\mathbf{j}\) and \(\mathbf{b} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) or equivalent	B1ft
	(5)
	(12 marks)

Notes

1B1 Equation of \(l_2\)

1M1 Using inverse transformation matrix correctly

2M1 Finding general point in terms of \(\lambda\).

A1 CAO for general point in terms of one parameter

2B1 ft for vector equation of their \(l_1\)