M1: Makes a start to the problem by finding the eigenvalues of \(\mathbf{A}\)
A1: Correct eigenvalues
M1: Uses a correct method to find an eigenvector for any eigenvalue
A1: One correct eigenvector (allow any integer multiple)
A1: Both correct eigenvectors (allow any integer multiples)
B1ft: For a matrix \(\mathbf{P}\) with their eigenvectors as columns
B1ft: For \(\mathbf{D}\) as a matrix with their eigenvalues on the leading diagonal. Must be consistent with their \(\mathbf{P}\) if \(\mathbf{P}\) is attempted.
Note method must be shown for full marks. Answer that obtain eigenvectors from a calculator can score M1A1 (if method shown for eigenvalues) M0A0A0 B1ftB1ft max.
(a) Determine the eigenvalues of matrix \(\mathbf{A}\) (3)
(b) Hence determine an orthogonal matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that\[\mathbf{D} = \mathbf{P}^{\mathrm{T}}\mathbf{A}\mathbf{P}\] (5)
Mark scheme (a)
Scheme
Marks
AO
\(\begin{vmatrix} 1-\lambda & -2 \\ -2 & 4-\lambda \end{vmatrix} = 0\) leading to \((1-\lambda)(4-\lambda) - 4 = 0\)
dM1: Dependent on the previous method mark. Solves their 3TQ to find a value for \(\lambda\)
A1: Correct values for \(\lambda\)
Mark scheme (b)
Scheme
Marks
AO
\(\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{their } \text{‘}0\text{’}\begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(x - 2y = 0 \Rightarrow x = 2y\) or \(\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{their } \text{‘}5\text{’}\begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(x - 2y = 5x \Rightarrow 2y = -4x\) or \(-2x + 4y = 5y \Rightarrow y = -2x\)
Note this appears on epen as M1A1A1A1A1 but is being marked as M1A1A1B1A1
M1: Uses \(\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{‘their } \lambda\text{’}\begin{pmatrix} x \\ y \end{pmatrix}\) to form an equation of the form \(ax = by\) for at least one of their eigenvalues. May be implied by a correct eigenvector.
A1: Deduces one correct eigenvector for one of the (correct) eigenvalues. Accept any non-zero multiple.
A1: Deduces both correct eigenvectors for the correct eigenvalues. Accept any non-zero multiples.
B1ft: Note: not dependent on the method mark. Deduces a correct matrix \(\mathbf{P}\) or \(\mathbf{D}\), following through on their eigenvalues or non-zero eigenvectors. This may be scored for a correct \(\mathbf{D}\) even if no attempt at the eigenvectors has been made, or a correct f.t. \(\mathbf{P}\) even if the method for eigenvectors was incorrect.
A1ft:Depends on the M having been scored. Correct matrices \(\mathbf{D}\) and \(\mathbf{P}\) which are consistent. Follow through on their eigenvalues and non-zero eigenvectors.
Note if they assume the eigenvector for \(\lambda = 0\) is 0 then do not allow this for the follow through mark(s) (though the first may be gained for correct D).
SC: If they mislabel P and D then allow A1ftA0ft for both correct but the wrong order, but A0A0 if only one is “correct” but wrong order.
\[\mathbf{A} = \begin{pmatrix} 4 & 2 & 0 \\ 2 & p & -2 \\ 0 & -2 & 2 \end{pmatrix} \qquad \text{where } p \text{ is a constant}\]
Given that \(\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) is an eigenvector of \(\mathbf{A}\),
(a) determine the eigenvalue corresponding to this eigenvector. (2)
(b) Hence show that \(p = 3\) (1)
(c) Determine
(i) the remaining eigenvalues of \(\mathbf{A}\),
(ii) corresponding eigenvectors for these eigenvalues. (6)
(d) Hence determine a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{\mathrm{T}}\) (3)
M1: Uses either the first or third row of matrix with the eigenvector to form and solve an equation in \(\lambda\). May see the full matrix equation used, but this is not necessary.
Alternative 1: multiples the matrix by the eigenvector and compares to a multiple of the eigenvector to deduce a value for \(\lambda\)
Alternative 2: Uses \(\begin{pmatrix} 4-\lambda & 2 & 0 \\ 2 & p-\lambda & -2 \\ 0 & -2 & 2-\lambda \end{pmatrix}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = 0\) to find an equation and find a value for \(\lambda\)
A1: For the correct eigenvalue of 3.
Mark scheme (b)
Scheme
Marks
AO
\(2 \times 2 - p - 2 \times 2 = 3 \times -1 \Rightarrow 4 - p - 4 = -3 \Rightarrow p = 3*\) Alternative 1 \((4-3)\left((p-3)(2-3) - 4\right) - 2\left(2(2-3) - 0\right) + 0 = 0 \Rightarrow p = 3*\) Using Alternative 2 from (a) \(4 - (p - \lambda) - 4 = 0\) uses \(\lambda = 3\) to show \(p = 3*\)
B1*
1.1b
(1)
Notes
B1*: Uses the second row with eigenvector 3 to set up a correct equation in \(p\) and then solves correctly.
Alternative 1: finds the characteristic equations, use \(\lambda = 3\) and correctly shows that \(p = 3\)
Alternative 2: Uses Alternative 2 from (a) and \(\lambda = 3\) to correctly show that \(p = 3\)
B1: Recalls that the eigenvalues form the entries of the diagonal matrix in the diagonalisation process. May be implied by working or stated, for work in (a) or (b). Implied by an attempt at equating the determinants of the two matrices.
M1: For a full method to find \(k\), e.g. by finding the characteristic equation and using (one of) the eigenvalues to find \(k\). The factor theorem can be used with either eigenvalue, for example, after finding the characteristic equation.
A1: Correct value.
Note Using \(k = -6\)
B1: Recalls that the eigenvalues form the entries of the diagonal matrix in the diagonalisation process.
M1: Uses \(k = -6\), finds the characteristic equation and solves to find the eigenvalues.
A1: Draws a conclusion, same eigenvalues therefore \(k = -6\)
Mark scheme (b)
Scheme
Marks
AO
For \(\lambda = -3\) eigenvector equations are \(\left\{\begin{matrix} 6x - \text{“}6\text{”}y = 0 \\ -5x + 5y = 0 \end{matrix}\right. \Rightarrow x, y = \ldots\) OR For \(\lambda = 8\) eigenvector equations are \(\left\{\begin{matrix} -5x - \text{“}6\text{”}y = 0 \\ -5x - 6y = 0 \end{matrix}\right. \Rightarrow x, y = \ldots\)
M1
2.1
(For \(\lambda = -3\), \(x = y\), for \(\lambda = 8\), \(5x = -6y\) so eigenvectors are) One of \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) o.e and \(\begin{pmatrix} 6 \\ -5 \end{pmatrix}\) o.e.
A1
1.1b
Both of \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) o.e. and \(\begin{pmatrix} 6 \\ -5 \end{pmatrix}\) o.e.
Given that matrix \(\mathbf{M}\) has a repeated eigenvalue,
(a) determine
(i) the value of \(k\)
(ii) the eigenvalue.
(6)
(b) Hence determine a Cartesian equation of the invariant line under \(T\). (2)
Mark scheme (a)(i)
Scheme
Marks
AO
\(\begin{vmatrix} 5-\lambda & 1 \\ k & -3-\lambda \end{vmatrix} = (5-\lambda)(-3-\lambda) - k = 0\)
M1
2.1
\(\lambda^2 - 2\lambda - k - 15 = 0\)
A1
1.1b
\(b^2 - 4ac = (-2)^2 - 4(1)(-k - 15) = 0 \Rightarrow k = \ldots\) or \(-k - 15 = 1 \Rightarrow k = \ldots\)
M1
1.1b
\(k = -16\)
A1
1.1b
(4)
Notes
M1: Starts the process by finding the determinant of \(\mathbf{M} - \lambda\mathbf{I}\) and sets = 0.
A1: Correct quadratic equation.
M1: Sets the discriminant of their 3TQ = 0 to find a value for \(k\). Alternatively identifies that for a perfect square the constant term must equal 1 and finds a value for \(k\).
M1: Uses their value of \(k\) to form and solve their 3TQ to find the eigenvalue. This mark can be implied by a correct eigenvalue.
A1: Correct eigenvalue
Mark scheme (b)
Scheme
Marks
AO
\(\begin{pmatrix} 5 & 1 \\ \text{their } k & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(ax + by = 0\) or \(ax = by\)
\(\begin{pmatrix} 5 & 1 \\ \text{their } k & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \text{their } \lambda\begin{pmatrix} x \\ y \end{pmatrix}\) leading to \(ax + by = 0\) or \(ax = by\)
Alternative approaches \(\begin{pmatrix} 5 & 1 \\ -16 & -3 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x \\ mx \end{pmatrix}\) leading to \(\left.\begin{matrix} 5x + mx = x \\ -16x - 3mx = mx \end{matrix}\right\}; {-16x} - 3mx = m(5x + mx)\) Leading to a value for \(m\) \(\{m^2 + 8m + 16 = 0\}\)
\(\begin{pmatrix} 5 & 1 \\ -16 & -3 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} x \\ mx + c \end{pmatrix}\) leading to \(\left.\begin{matrix} 5x + mx + c = x \\ -16x - 3mx - 3c = mx + c \end{matrix}\right\}; {-16x} - 3mx - 3c = m(5x + mx + c) + c\) Leading to a value for \(m\) \(\{-3c = mc + c\ \text{ or }\ m^2 + 8m + 16 = 0\}\)
M1
2.1
\(y = -4x\) o.e.
A1
1.1b
(2)
(8 marks)
Notes
M1: Constructs a rigorous argument using their eigenvalue to find the Cartesian equation of the invariant line.
(a) Determine, in expanded form in terms of \(a\), the characteristic equation for \(\mathbf{A}\). (2)
(b) Hence use the Cayley-Hamilton theorem to determine values of \(a\) and \(b\) such that\[\mathbf{A}^3 = \mathbf{A} + b\mathbf{I}\]where \(\mathbf{I}\) is the \(2 \times 2\) identity matrix. (4)
\(\Rightarrow \mathbf{A}^3 = (3a + 8 + 49)\mathbf{A} + 7(3a + 8)\mathbf{I} \Rightarrow 3a + 57 = 1 \Rightarrow a = \ldots\) Or \(\begin{pmatrix} b-1 & a \\ 3 & b+8 \end{pmatrix} = \begin{pmatrix} 18a-1 & 3a^2+57a \\ 171+9a & 512+45a \end{pmatrix} \Rightarrow\) e.g. \(3 = 171 + 9a \Rightarrow a = \ldots\) Or \(\mathbf{A}^3 = \begin{pmatrix} -1 & a \\ 3 & 8 \end{pmatrix} + \begin{pmatrix} 18a & 3a^2+56a \\ 168+9a & 504+45a \end{pmatrix} \Rightarrow\) e.g. \(18a = 504 + 45a \Rightarrow a = \ldots\)
M1
1.1b
\(\Rightarrow a = -\dfrac{56}{3},\ b = -336\)
A1
1.1b
(4)
(6 marks)
Notes
Note: this question asks the candidates to use the Cayley-Hamilton theorem so any other approach that doesn’t score the first method mark scores no marks
M1: Uses the Cayley-Hamilton theorem with their equation and multiplies though by \(\mathbf{A}\) to find an equation for \(\mathbf{A}^3\)
M1: Substitutes for \(\mathbf{A}^2\) to obtain an equation for \(\mathbf{A}^3\) in terms of \(a\), \(\mathbf{I}\) and \(\mathbf{A}\). Alternatively substitutes in the matrix \(\mathbf{A}\) and attempts to square
M1: Equates coefficient(s) of \(\mathbf{A}\) to 1 and proceeds to find a value for \(a\). Alternatively equates elements to find a value for \(a\).
A1: Correct values for \(a\) and \(b\).
Special case: Missing matrix \(\mathbf{I}\) from their working can score maximum of M1 (if multiply by A correctly) M1M1A0 unless implied \(\mathbf{I}\) from their working this can score all marks.
(corrected from the printed mark scheme: in the two “Or” methods for the third mark the bottom-right entries are printed as \(512 + 27a\) and \(504 + 27a\), and the example as \(18a = 504 + 27a\); the correct entries are \(512 + 45a\) and \(504 + 45a\))
\(\lambda^2 \Rightarrow\ a+b+1 = 7\) and \(\lambda \Rightarrow\ a+b+ab-1 = 13\) Solves simultaneously e.g. \(a+b = 6,\ ab = 8\) For example: leading to \(a^2 - 6a + 8 = 0 \Rightarrow a = \ldots\)
M1
3.1a
\(a = 2,\ b = 4\)
A1
1.1b
\(c = -1\)
A1
2.2a
(5)
Notes
M1: Correct method to find the characteristic equation for \(\mathbf{M}\), condone missing = 0, and one slip as long as the intention is clear
A1: Multiplies out to achieve a correct characteristic equation, condone missing = 0
M1: A complete method to find the values of the constants \(a\) or \(b\). Equates their coefficients for \(\lambda^2\) and \(\lambda\) and solves simultaneously to find values for \(a\) or \(b\).
A1: Deduces the correct values for \(a\) and \(b\). \((a \lt b)\) following correct simultaneous equations
B1ft: Uses Cayley-Hamilton theorem to produce equation replacing \(\lambda\) with \(\mathbf{M}\) and constant term with constant multiple of the identity matrix \(\mathbf{I}\). Follow through on their value for \(c\). This mark may be implied by the M mark.
M1: A complete method to find \(M^{-1}\) using the Cayley-Hamilton theorem. The minimum is for writing an expression for \(M^{-1}\) from their characteristic equation, for example \(M^{-1} = M^2 - 7M + 13I\) and then stating an answer for \(M^{-1}\), they may have used their calculator, there is no need to check.
and \(\mathbf{D} = \begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}\) or \(\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix}\) Note: If \(\mathbf{P}\) is given, their \(\mathbf{D}\) must be consistent with it to award this mark.
B1ft
2.2a
(7)
(7 marks)
Notes
M1: Begins the process of finding suitable matrices by attempting the eigenvalues of \(\mathbf{M}\).
A1: Correct eigenvalues.
M1: Correct method to find an eigenvector for either of their eigenvalues.
A1: One correct eigenvector – accept any non-zero multiples of their eigenvectors.
A1: A correct eigenvector for each of their eigenvalues.
B1ft: Gives \(\mathbf{P}\) as a matrix with their columns as the eigenvectors found.
B1ft: Gives \(\mathbf{D}\) as the matrix with eigenvalues on the diagonal – which must be in the correct order for their \(\mathbf{P}\) (if it is given).
\[\mathbf{A} = \begin{pmatrix} 5 & -2 & 5 \\ 0 & 3 & p \\ -6 & 6 & -4 \end{pmatrix} \qquad \text{where } p \text{ is a constant}\]
Given that \(\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}\) is an eigenvector for \(\mathbf{A}\)
(a)
(i) determine the eigenvalue corresponding to this eigenvector (1)
(ii) hence show that \(p = 2\) (2)
(iii) determine the remaining eigenvalues and corresponding eigenvectors of \(\mathbf{A}\) (7)
(b) Write down a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A} = \mathbf{PDP}^{-1}\) (1)
(c)
(i) Solve the differential equation \(\dot{u} = ku\), where \(k\) is a constant. (2)
With respect to a fixed origin \(O\), the velocity of a particle moving through space is modelled by
\[\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \mathbf{A}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\]
By considering \(\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \mathbf{P}^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\) so that \(\begin{pmatrix} \dot{u} \\ \dot{v} \\ \dot{w} \end{pmatrix} = \mathbf{P}^{-1}\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix}\)
(ii) determine a general solution for the displacement of the particle. (4)
M1: Applies \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) to achieve a cubic in \(\lambda\) (or other variable, simplification not required). Allow with \(p\) used instead of 2, and look for two correct “terms” in the expansion leading to a cubic as evidence of the expansion.
A1: Correct simplified cubic. Note this may be implied by correct answers from a calculator following a correct expansion seen for the M.
A1: Correct eigenvalues
M1: Forms and solves eigenvector equations for at least one (other than \(-1\)) eigenvalue.
B1ft: A correct corresponding P and D, follow through on their answer to (a). Columns may be in different order, but should be consistent for their P and D.
Mark scheme (c)(i)
Scheme
Marks
AO
\(\dot{u} = ku \Rightarrow \displaystyle\int \frac{1}{u}\,\mathrm{d}u = k\int \mathrm{d}t \Rightarrow \ln u = kt (+c)\)
M1
1.1b
So \(u = A\mathrm{e}^{kt}\) or \(u = \mathrm{e}^{kt + c}\)
A1
1.1b
(2)
Notes
M1: Separates variables and attempts the integration (constant not required).
A1: Correct answer for \(u = \ldots\), either form, including constant of integration
Mark scheme (c)(ii)
Scheme
Marks
AO
\(\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \mathbf{PDP}^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix} \Rightarrow \begin{pmatrix} \dot{u} \\ \dot{v} \\ \dot{w} \end{pmatrix} = \mathbf{P}^{-1}\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \mathbf{D}\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} -u \\ 2v \\ 3w \end{pmatrix}\)
M1
3.1b
\(\Rightarrow \begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} A\mathrm{e}^{-t} \\ B\mathrm{e}^{2t} \\ C\mathrm{e}^{3t} \end{pmatrix}\)
M1
2.2a
\(\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{P}\begin{pmatrix} A\mathrm{e}^{-t} \\ B\mathrm{e}^{2t} \\ C\mathrm{e}^{3t} \end{pmatrix} = \ldots\)
M1
3.4
\(\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2A\mathrm{e}^{-t} + 3B\mathrm{e}^{2t} + C\mathrm{e}^{3t} \\ A\mathrm{e}^{-t} + 2B\mathrm{e}^{2t} + C\mathrm{e}^{3t} \\ -2A\mathrm{e}^{-t} - B\mathrm{e}^{2t} \end{pmatrix}\)
A1
1.1b
(4)
(17 marks)
Notes
NB different orderings of the columns of P and D will give the terms in different orders here.
M1: Uses their P and D to transform system into equation in \(u\), \(v\) and \(w\) (may be implied).
M1: Forms the solution for \(u\), \(v\) and \(w\) using their eigenvalues.
M1: Reverses the substitution (multiplies by their P) to get solution for \(x\), \(y\) and \(z\).
A1: Correct answer, in matrix form or as separate equations – award when first seen and isw.
M1: Attempts to use determinant equals 5 to find \(k\). May be attempted by finding determinant from original matrix, or attempt at using the \(\text{“}{-5}(k + 6)\text{”}\) from the expansion in (a) (allow \(\pm\) for the method mark).
A1: \(k = -7\)
(ii)
M1: Attempts to use the Cayley-Hamilton theorem to set up a matrix equation. The equation should be correct for their \(k\), including correct use of \(\mathbf{I}\).
M1: Realises the need to multiply the equation through (either side) by \(\mathbf{M}^{-1}\) and rearrange to make \(\mathbf{M}^{-1}\) the subject.
(a) Show that the characteristic equation for \(\mathbf{A}\) is \(\lambda^2 - 5\lambda + 6 = 0\) (2)
(b) Use the Cayley-Hamilton theorem to find integers \(p\) and \(q\) such that\[\mathbf{A}^3 = p\mathbf{A} + q\mathbf{I}\] (3)
(ii) Given that the \(2 \times 2\) matrix \(\mathbf{M}\) has eigenvalues \(-1 + \mathrm{i}\) and \(-1 - \mathrm{i}\), with eigenvectors \(\begin{pmatrix} 1 \\ 2 - \mathrm{i} \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 2 + \mathrm{i} \end{pmatrix}\) respectively, find the matrix \(\mathbf{M}\). (5)
M1: A full method leading to \(\lambda^3\) in terms of \(\lambda\)
M1: Applies the Cayley-Hamilton theorem
A1: Deduces the correct expression or correct values for \(p\) and \(q\)
Mark scheme (ii)
Scheme
Marks
AO
\(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 2-\mathrm{i} \end{pmatrix} = (-1+\mathrm{i})\begin{pmatrix} 1 \\ 2-\mathrm{i} \end{pmatrix}\) or \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 2+\mathrm{i} \end{pmatrix} = (-1-\mathrm{i})\begin{pmatrix} 1 \\ 2+\mathrm{i} \end{pmatrix}\)
\(a + b(2-\mathrm{i}) = -1+\mathrm{i},\ a + b(2+\mathrm{i}) = -1-\mathrm{i} \Rightarrow a = 1,\ b = -1\) or \(c + d(2-\mathrm{i}) = -1+3\mathrm{i},\ c + d(2+\mathrm{i}) = -1-3\mathrm{i} \Rightarrow c = 5,\ d = -3\)
M1: Attempts to expand the determinant to find the characteristic polynomial.
Note: other methods of expanding the determinant are possible. If unsure send to review.
A1: Correct expansion need not be simplified. (Need not see set equal to zero) Allow recovery of missing brackets if indicated by later working.
M1: Attempts to take out a factor of \((\lambda - 2)\) of their equation (may first expand to cubic or may spot the factor and take out without full expansion). E.g \((6 - \lambda)\left((3 - \lambda)^2 - 1\right) + 2\big(2(\lambda - 3) + 2\big) + 2\big(2 - 2(3 - \lambda)\big) = (6 - \lambda)(4 - \lambda)(2 - \lambda) + 4(\lambda - 2) + 4(\lambda - 2)\) \(= (\lambda - 2)\big({-(6 - \lambda)(4 - \lambda)} + 4 + 4\big)\)
(corrected from the printed mark scheme: the last line is printed as \((\lambda - 2)\big((6 - \lambda)(4 - \lambda) + 4 + 4\big)\), without the minus sign)
This is for a method that will allow \(\lambda\) to be shown as a repeated eigenvalue, so just stating two solutions is not sufficient, factorisation must be seen.
A1*: Obtains a correct factor of \((\lambda - 2)^2\) and deduces that 2 is a repeated eigenvalue. Must see statement about 2 being repeated. (Just listing 2 twice is not sufficient.)
B1: (Note this is A1 on ePEN) Obtains and identifies 8 as the other eigenvalue (B0 if not identified in (a) but full marks can be scored in (b) and (c) for use of 8 as eigenvalue)
Mark scheme (b)
Scheme
Marks
AO
\(\begin{pmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{pmatrix}\mathbf{v} = 2\begin{pmatrix} x \\ y \\ z \end{pmatrix}\) or \(\begin{pmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{pmatrix}\mathbf{v} = 8\begin{pmatrix} x \\ y \\ z \end{pmatrix} \Rightarrow \mathbf{v} = \ldots\)
M1
1.1b
Obtains any multiple of \(\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) for \(\lambda = 8\)
A1
1.1b
Obtains any (non-zero) multiple or linear combination of \(\begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}\text{ or }\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\text{ or }\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) for \(\lambda = 2\)
A1
1.1b
Obtains a different linear combination or (non-zero) multiple of different vector from \(\begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}\text{ or }\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\text{ or }\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) for \(\lambda = 2\)
A1
3.1a
(4)
Notes
M1: Uses a correct method to find at least one eigenvector
A1: Obtains one correct eigenvector for \(\lambda = 8\)
A1: Obtains one correct eigenvector for \(\lambda = 2\)
A1: Obtains two correct linearly independent eigenvectors for \(\lambda = 2\)
Note some other common eigenvectors for \(\lambda = 2\) are \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}\)
M1: Forms a matrix with their three different non-zero eigenvectors as columns or with their normalised (or any scaled version) of their eigenvectors.
A1ft: Correct matrix with the eigenvectors (normalised/scaled) as columns in any order (follow through their three different vectors which are not multiples of any other)
(a) find the characteristic equation for the matrix \(\mathbf{A}\), simplifying your answer. (2)
(b) Hence find an expression for the matrix \(\mathbf{A}^{-1}\) in the form \(\lambda\mathbf{A} + \mu\mathbf{I}\), where \(\lambda\) and \(\mu\) are constants to be found. (3)
M1: Complete method to find the characteristic equation, condone missing = 0
A1: Obtains a correct three term quadratic equation – may use any variable.
Mark scheme (b)
Scheme
Marks
AO
\(\mathbf{A}^2 - 5\mathbf{A} + 2\mathbf{I} = 0\)
B1ft
1.1b
Multiplies through by \(\mathbf{A}^{-1}\) \(\mathbf{A} - 5\mathbf{I} + 2\mathbf{A}^{-1} = 0\) and rearranges to get \(\mathbf{A}^{-1} = \ldots\) OR Rearranges to make \(\mathbf{I}\) the subject, takes out a factor of \(\mathbf{A}\) and rearranges to get \(\mathbf{A}^{-1} = \ldots\) \(\mathbf{I} = \dfrac{(5\mathbf{A} - \mathbf{A}^2)}{2} = \mathbf{A}\dfrac{(5\mathbf{I} - \mathbf{A})}{2} \Rightarrow \mathbf{A}^{-1} = \ldots\) OR Rearranges to make \(\mathbf{I}\) the subject and multiplies through by \(\mathbf{A}^{-1}\) \(\mathbf{I} = \dfrac{5}{2}\mathbf{A} - \dfrac{1}{2}\mathbf{A}^2 \Rightarrow \mathbf{A}^{-1} = \dfrac{5}{2}\mathbf{A}\mathbf{A}^{-1} - \dfrac{1}{2}\mathbf{A}^2\mathbf{A}^{-1}\)
B1ft: Uses Cayley Hamilton Theorem to produce equation replacing \(\lambda\) with \(\mathbf{A}\) and constant term with constant multiple of the identity matrix \(\mathbf{I}\)
M1: A complete method using part (a) to find \(\mathbf{A}^{-1}\) Multiplies through by \(\mathbf{A}^{-1}\) and rearranges to get \(\mathbf{A}^{-1} = \ldots\) Or rearranges to make \(\mathbf{I}\) the subject, takes out a factor of \(\mathbf{A}\), and rearranges to get \(\mathbf{A}^{-1} = \ldots\) Or rearranges to make \(\mathbf{I}\) the subject and multiplies through by \(\mathbf{A}^{-1}\) to get \(\mathbf{A}^{-1} = \ldots\)
A1: Correct expression for \(\mathbf{A}^{-1}\), must be using their answer to part (a).
\(\begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 2\begin{pmatrix} x \\ y \end{pmatrix}\) or \(\begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 3\begin{pmatrix} x \\ y \end{pmatrix}\)
