FP3 June 2014 Q2

EdexcelOld spec13 marksMatrices

2. \[\mathbf{M} = \begin{pmatrix}1 & 0 & 2 \\ 0 & 4 & 1 \\ 0 & 5 & 0\end{pmatrix}\]

(a) Show that matrix \(\mathbf{M}\) is not orthogonal. (2)

(b) Using algebra, show that 1 is an eigenvalue of \(\mathbf{M}\) and find the other two eigenvalues of \(\mathbf{M}\). (5)

The transformation \(M : \mathbb{R}^3 \to \mathbb{R}^3\) is represented by the matrix \(\mathbf{M}\).

(d) Find a cartesian equation of the image, under this transformation, of the line \[x = \frac{y}{2} = \frac{z}{-1}\] (4)

Mark scheme (a)

Scheme	Marks
\(\mathbf{M}\mathbf{M}^{\mathrm{T}} = \begin{pmatrix}1 & 0 & 2 \\ 0 & 4 & 1 \\ 0 & 5 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 4 & 5 \\ 2 & 1 & 0\end{pmatrix}\) Attempts \(\mathbf{M}\mathbf{M}^{\mathrm{T}}\) or \(\mathbf{M}^{\mathrm{T}}\mathbf{M}\) or scalar product of at least one pair of columns or attempts magnitude of at least one column or finds \(\det\mathbf{M}\) or attempts \(\mathbf{M}^{-1}\)	M1
\(= \begin{pmatrix}5 & 2 & 0 \\ 2 & 17 & 20 \\ 0 & 20 & 25\end{pmatrix} \neq \mathbf{I}\) \(\therefore \mathbf{M}\) not orthogonal. or scalar product \(\neq 0\) or magnitude \(\neq 1\) or \(\det\mathbf{M} \neq \pm 1\) (must see \(\pm\)) or \(\mathbf{M}^{-1} \neq \mathbf{M}^{\mathrm{T}}\) and conclusion. Note that not all of \(\mathbf{M}\mathbf{M}^{\mathrm{T}}\) or \(\mathbf{M}^{-1}\) is necessary and there may be errors but there must be some correct work (at least one correct relevant element). NB \(\det\mathbf{M} = -5\). See extra notes for \(\mathbf{M}^{-1}\)	A1
	(2)

Notes

Extra Notes

\(\mathbf{M}^{-1} = \dfrac{1}{5}\begin{pmatrix}5 & -10 & 8 \\ 0 & 0 & 1 \\ 0 & 5 & -4\end{pmatrix}\) \(\mathbf{M}^{\mathrm{T}}\mathbf{M} = \begin{pmatrix}1 & 0 & 2 \\ 0 & 41 & 4 \\ 2 & 4 & 5\end{pmatrix}\)

Mark scheme (b)

Scheme	Marks
\(\begin{vmatrix}1 - \lambda & 0 & 2 \\ 0 & 4 - \lambda & 1 \\ 0 & 5 & -\lambda\end{vmatrix} = 0\) This statement is sufficient. (allow other brackets provided the determinant is implied later)	M1
\((1 - \lambda)\left[(4 - \lambda)(-\lambda) - 5\right] - 0(0 - 0) + 2(0 - 0) = 0\) Attempts characteristic equation (= 0 may be implied by their value(s) for \(\lambda\)) Allow one slip e.g. – usually the omission of the “\(-5\)”	M1
\((1 - \lambda)((4 - \lambda)(-\lambda) - 5) = 0\)
\(\lambda = 1\) \(\lambda = 1\) with no errors	A1cso
\(\lambda^2 - 4\lambda - 5 = 0\) \(\Rightarrow \lambda = 5, \lambda = -1\) M1: Attempts to find the other 2 eigenvalues from their characteristic equation by solving a 3 term quadratic. A1: \(\lambda = 5, \lambda = -1\)	M1A1
	(5)

Notes

(corrected from the printed mark scheme: the second line is printed as \((1 - \lambda)\left[(4 - \lambda)(-\lambda) - 5\right]\left(-0(0 - 0) + 2(0 - 0)\right) = 0\), with the last two terms bracketed as a product; they are added terms of the determinant expansion)

Mark scheme (c)

Scheme	Marks
\(\begin{pmatrix}1 & 0 & 2 \\ 0 & 4 & 1 \\ 0 & 5 & 0\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}x \\ y \\ z\end{pmatrix}\) or \(\begin{pmatrix}0 & 0 & 2 \\ 0 & 3 & 1 \\ 0 & 5 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}\) A correct statement for the eigenvalue 1. (May be implied by correct equations)	M1
\(\alpha(\mathbf{i} + 0\mathbf{j} + 0\mathbf{k})\) where \(\alpha\) is a constant Any vector of this form.	A1
	(2)

Mark scheme (d)

Scheme	Marks
\(\begin{pmatrix}1 & 0 & 2 \\ 0 & 4 & 1 \\ 0 & 5 & 0\end{pmatrix}\begin{pmatrix}t \\ 2t \\ -t\end{pmatrix} = \begin{pmatrix}-t \\ 7t \\ 10t\end{pmatrix}\) M1: Attempt to multiply the parametric form or direction of the line by M. Condone use of 0 for the \(x\) component but the line must pass through the origin. A1: Correct image vector with or without “\(t\)”	M1A1
Cartesian equation \(\dfrac{x}{-1} = \dfrac{y}{7} = \dfrac{z}{10}\) M1: Correct method to convert to cartesian form of a straight line passing through the origin. A1: Correct equations (any multiple)	M1A1
	(4)
	(13 marks)