Notes

M1: Finds the dot product of the normal vectors, sets equal to 0 and finds a value for \(a\). Condone a sign slip.

A1: Correct value for \(a\)

Notes

M1: Finds the determinant of a 3 by 3 matrix and sets = 0, this may be implied

A1: Correct determinant

A1: Correct value for \(b\)

dM1: Uses the equations of the three planes to find two equations by eliminating the same variable. Condone a slip on one coefficient or constant term

ddM1: Correctly uses their equations to form and solve an equation for \(c\)

A1: Correct value for \(c\)

Alternative 1

M1: Uses the planes to eliminate one variable to form two equations

A1: One correct equation

A1: two correct equations

dM1: Sets up an equation for \(b\) or \(c\) by comparing coefficients and solves to find a value. Condone a slip on one coefficient or constant term

ddM1: Solves equations to find values for \(b\) and \(c\)

A1: Correct value for \(b\) and \(c\) cso

Note for 10 (b) using Alternative 1

Alternative 2

M1: Uses the planes to eliminate one variable to form two equations or substitutes in two values for one coordinate and solves to find two coordinates that lie on the line of intersection

A1: One correct equation or coordinate

A1: Two correct equations or coordinates

dM1: \(\begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) leading to \(8 - 7b + 13 = 0 \Rightarrow b = \ldots\) or uses their coordinates to find the direction vector and uses that it is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) to find a value for \(b\)

ddM1: \((5, -2, 0)\) lies on \(\Pi_3\) leading to \(5 - 2b = c\)

A1: Correct values for \(b\) and \(c\)

Notes

M1: Finds the determinant of the matrix, sets \(= 0\) and solves to find a value of \(p\). Allow for any recognisable attempt at finding the determinant (may be slips in coefficients or signs). Accept if det \(= 14 - 7p\) appears with no working shown.

A1: \(p = 2\) from correct work.

Alt I Using normal

M1: Attempts the cross product between sets of pairs of planes and scales (if appropriate) and equates directions to solve for \(p\).

A1: Correct \(p\)

FYI
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -1 & 1\\ 1 & p & -3\end{vmatrix} = (3 - p)\mathbf{i} + 7\mathbf{j} + (2p + 1)\mathbf{k},\quad \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -1 & 1\\ 3 & 1 & -2\end{vmatrix} = \mathbf{i} + 7\mathbf{j} + 5\mathbf{k},\quad \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & p & -3\\ 3 & 1 & -2\end{vmatrix} = (3 - 2p)\mathbf{i} - 7\mathbf{j} + (1 - 3p)\mathbf{k}\)

Notes

M1: A complete method to find the value of \(q\). May be implied by correct value if no incorrect working is shown.
E.g. Adds together equations 1 and 2 and compares with equation 3 to find a value for \(q\).
Alternatively: Uses equations \(2x - y + z = 3\) and \(3x + y - 2z = 4\) to eliminate one variable. Uses equation \(x + 2y - 3z = q\) and one of the other equations to eliminate the same variable and compare to find a value for \(q\). Another possible method is to identify a point on both planes 1 and 3 (e.g. let \(x = 1\) and solve for \(y\) and \(z\) to get \((1,\ -3,\ -2)\) then substitute into middle equation to find \(q\))

A1: \(q = 1\)

Alternatively II the question may be done as a whole. Eg.
\(\left.\begin{aligned}2x - y + z &= 3\\ 3x + y - 2z &= 4\end{aligned}\right\} \Rightarrow x + 2y - 3z = 1 \Rightarrow p = 2,\ q = 1\) is the most direct method. Variations are possible, e.g.
\(\left.\begin{aligned}2x - y + z &= 3\\ x + py - 3z &= q\\ 3x + y - 2z &= 4\end{aligned}\right\} \Rightarrow \begin{aligned}&3(1) + (2): 7x + (p - 3)y = 9 + q\\ &2(1) + (3): \quad\ 7x - y = 10\end{aligned} \Rightarrow \begin{aligned}&p - 3 = -1 \Rightarrow p = \ldots\\ &9 + q = 10 \Rightarrow q = \ldots\end{aligned}\)

Score as follows:

M1: Attempts to solve at least two of the equations simultaneously to eliminate one variable or match coefficients of the third equation and uses the equation to identify one of the unknowns. Condone slips as long as the method is clear.

A1: Correct \(p\) or correct \(q\).

M1: Full method to use the linear dependence of the equations to find both variables.

A1: Correct \(p\) and \(q\).

Alt III: Using points on the common line.

M1: Finds two separate points on each of the first and third plane.

A1: Two correct points.

M1: Forms and solves two equations in \(p\) and \(q\) using their points.

A1: Correct \(p\) and \(q\).

E.g.
\(x = 0 \Rightarrow \begin{cases}-y + z = 3\\ y - 2z = 4\end{cases} \Rightarrow y = -10,\ z = -7\)
\(x = 1 \Rightarrow \begin{cases}-y + z = 1\\ y - 2z = 1\end{cases} \Rightarrow y = -3,\ z = -2\)
\(\Rightarrow \begin{aligned}-10p + 21 &= q\\ 1 - 3p + 6 &= q\end{aligned} \Rightarrow p = \ldots,\ q = \ldots\)

Some useful equations:

Eliminating \(x\): \(\begin{cases}(2p + 1)y - 7z = 2q - 3\\ 5y - 7z = -1\\ (3p - 1)y - 7z = 3q - 4\end{cases}\)

Eliminating \(y\): \(\begin{cases}(2p + 1)x + (p - 3)z = 3p + q\\ 5x - z = 7\\ (3p - 1)x + (3 - 2p)z = 4p - q\end{cases}\)

Eliminating \(z\): \(\begin{cases}7x + (p - 3)y = 9 + q\\ 7x - y = 10\\ 7x + (3 - 2p)y = 12 - 2q\end{cases}\)

Notes

B1: See scheme, enlargement is B0

B1: See scheme, if they mention of \(y\)-axis must be correct e.g. scale factor 1, stays the same.
Condone defining the stretch as from/about the origin and in the \(x\)- axis, along the \(x\)-axis

Notes

B1: See scheme, any incorrect answer stated B0

Notes

M1: Correct method to find the area of the triangle \(T\)

B1: Correct determinant of the matrix

M1: Sets their determinant = 4.5 divided by their area of the triangle \(T\)

dM1: Dependent on previous method mark. Uses the identity \(\tan 2\theta = \dfrac{\sin 2\theta}{\cos 2\theta}\) to achieve \(\sin 2\theta = k\) where \(-1 \lt k \lt 1\)

ddM1: Dependent on previous method marks. Solves \(\sin 2\theta = k\) where \(-1 \lt k \lt 1\) to find a value of \(\theta\)

A1: Any two correct values in degrees or radians

A1: All eight correct values and no others must be in degrees

Notes

B1: Correct determinant

M1: Correct method to find the inverse matrix using their determinant

A1: Fully correct inverse matrix, isw

Notes

M1: A complete method to form two simultaneous equations in \(a\) and \(b\), using \(\mathbf{M} + \mathbf{M}^{-1} = \mathbf{I}\) and their inverse matrix

dM1: Solves their two different equations to achieve a value for \(a\) and a value for \(b\). Must come from a valid attempt at \(\mathbf{M} + \mathbf{M}^{-1} = \mathbf{I}\).

A1: Correct values for \(a\) and \(b\)

Alternative

(Corrected from the printed mark scheme: the bottom-right entry of \(\mathbf{M}^2\) is printed as \(-b + a\); it is \(-b + 1\), as used in the next line.)

M1: Forms the equation \(\mathbf{M}^2 + \mathbf{I} = \mathbf{M}\) and form two equations

dM1: Solves their equations to achieve a value for \(a\) and a value for \(b\)

A1: Correct values for \(a\) and \(b\)

Notes

B1: Shows that the result holds for \(n = 1\). Must see substitution in the RHS minimum required is \(\begin{pmatrix}1 & 5(2 - 1)\\ 0 & 2\end{pmatrix}\) and reaches \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\) No need to state “true for \(n = 1\)” for this mark.

M1: Assumes the result is true for some value of \(n = k\), and sets up a matrix multiplication of the assumed result multiplied by the original matrix, either way round (no need to carry out for this mark, it is for essentially explaining the procedure). “Assume (true for) \(n = k\)” (oe) is sufficient for the assumption and may even be part of the conclusion and accept any alternative wording that indicates the assumption has been made. Allow for setting up the multiplication in reverse (ie working from \(n = k + 1\) towards \(n = k\), \(\mathbf{A}^k = \mathbf{A}^{k+1}\mathbf{A}^{-1}\) oe)

M1: Carries out the multiplication. (Allow the reverse case.)

A1: Achieves a correct un-simplified matrix. Accept in working in reverse.

A1: Reaches a correct simplified matrix with no errors, the correct un-simplified matrix seen previously and at least one intermediate line which must be correct. If working from both sides all steps in showing the two sides are equal must be seen. If working in reverse they must return to \(\mathbf{A}^{k+1} = \ldots\) for this mark.

A1: Correct formal conclusion. This mark is dependent on the M and previous A marks having been scored and an attempt at the check for \(n = 1\) (if e.g. they didn’t show sufficient detail). It is gained by conveying the ideas of all three bold points at the end of their solution.

Note: Some cases may use \(n = 0\) as the base case. These can score full marks if dealt with correctly, but the conclusion must be consistent with their initial check to score the final A.
For the B mark minimum \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^0 = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) ( or \(\mathbf{I}\)) and \(\begin{pmatrix}1 & 5(1 - 1)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) should be seen. (No need to state true for \(n = 0\) for this mark – but must have consistent conclusion for the final A as noted above). If unsure send to review.

Notes

B1: Identifies the transformation as a reflection. Accept stretch with scale factor \(-1\).

B1: Identifies reflection and the correct line of reflection. Condone phrasing such as “across the \(y\)-axis”. Must be a single transformation – B0 if they state something else also happens.
NB Allow B1B0 for answers such as “flip in the y-axis” that convey the correct transformation in imprecise language.

Notes

B1: Correct matrix

Notes

M1: A complete method to find a value for \(2^n\) (or \(n\)). Multiplies the coordinates \((27,\ 1)\) by their matrix \(\mathbf{C}\) and sets the \(x\) and \(y\) coordinates equal to reach a value for \(2^n\) or \(n\) (condone negative values for \(2^n\)). Note you may allow this for reach a value of \(a\) if \(a = 2^n\) is stated or clearly implied by their working.

M1: Uses their value of \(2^n\) or \(n\), to find the matrix \(\mathbf{A}^n\). Note substituting into \(\mathbf{C}\) is M0 without further work to find \(\mathbf{A}^n\).

A1: Correct matrix.

Notes

M1: Finds the determinant of the matrix \(\mathbf{M}\) and sets \(= 0\)
Allow for sight of \(\pm 3(3a - a) \pm 6(a^2 - 2) \pm 0(a^2 - 6) = 0\) and do not allow sign slips in their minors.
Need not be simplified

dM1: Solves their 3TQ. Usual rules apply for solving a quadratic by any means including using a calculator.

A1: Both correct values for \(a\).

Note: Correct answers without a method can be awarded M1dM1A1

Alternatively, the determinant can be found using the vector product, for example:

If you are uncertain, please send to review

Notes

M1: A complete method to find a value for \(a\). Sets their determinant equal to \(-108\), solves a 3TQ and proceeds to find at least one negative value of \(a\).
They must use their determinant equal to \(-108\) and not for example a changed quadratic from (a) equal to \(-108\). Usual rules apply for solving a quadratic by any means including using a calculator.

A1: Deduces \(a = -4\) only, which could be implied by a correct matrix.

A1: Correct matrix. Accept exact equivalents but not answers rounded to decimal places. Isw once a correct matrix is seen.

e.g. Accept \(\dfrac{1}{108}\begin{pmatrix}8 & -24 & -6\\ 14 & 12 & 3\\ -10 & -24 & -33\end{pmatrix}\) or \(-\dfrac{1}{108}\begin{pmatrix}-8 & 24 & 6\\ -14 & -12 & -3\\ 10 & 24 & 33\end{pmatrix}\)

Notes

If no attempt at part (a) has been made you may award marks for work seen in part (b) for finding the inverse.

B1(M1 on epen): Correct determinant of \(2k\)

M1: Starts the process of finding the inverse and obtains at least 6 correct elements of cofactors
Alternatively transposes their matrix of minors and obtains at least 6 correct elements.

dM1: A complete recognisable method to find the inverse including dividing by the determinant
Allow minor slips if the process is clearly correct.

A1: Correct inverse.

Notes

M1: Attempts \(\mathbf{A}^{-1}\begin{pmatrix}3\\ 1\\ 6\end{pmatrix}\) with their \(\mathbf{A}^{-1}\) which must be in terms of \(k\), to obtain at least one of \(x =\), \(y =\) or \(z =\) which may be seen embedded in a column vector, simplified or unsimplified, the determinant may be outside their column vector.
Condone a slip in copying \(\begin{pmatrix}3\\ 1\\ 6\end{pmatrix}\) e.g. \(\begin{pmatrix}1\\ 3\\ 6\end{pmatrix}\) or \(\begin{pmatrix}3\\ 1\\ 2\end{pmatrix}\)

A1(M1 on epen): Two correct expressions for \(x\), \(y\) or \(z\) simplified or unsimplified, which may appear in a column vector, determinant cannot be outside the vector.

A1: Correct coordinates in simplest form. Allow e.g. \(y = 2 - \dfrac{12}{k}\)
Their final answer must be written as coordinates and not as a column vector but can be written as \(x =\), \(y =\) and \(z =\)

Alternative (using algebraic method for simultaneous equations):

M1: Solves simultaneously to obtain at least one of \(x =\), \(y =\) or \(z =\) must be in terms of \(k\).

A1(M1 on epen): Two correct expressions for \(x\), \(y\) or \(z\) simplified or unsimplified.

A1: Correct coordinates in simplest form. Allow e.g. \(y = 2 - \dfrac{12}{k}\)
Their final answer must be written as coordinates and not as a column vector but can be written as \(x =\), \(y =\) and \(z =\)

e.g.
Eliminates \(z\) and achieves \(4x + 2y = 4\) and \((6 - k)x + 3y = 0\)
Uses \(12x + 6y = 12\) and \((12 - 2k)x + 6y = 0\) to produce \(2kx = 12 \Rightarrow x = \dfrac{6}{k}\)

Notes

M1: Calculates one of the elements of the leading diagonal of AB, condone sign slips

A1: Sets diagonal = \(3k + c\) and deduces the correct value for \(c\). Award for sight of \(3k + 1\)

Notes

M1: Attempts to solve \(3k + \text{“}1\text{”} = 0\) or attempts the determinant, condone sign slips in the minors, and sets = 0 leading to a value for \(k\)

A1ft: Correct value or follow through their value for \(c\) so allow for \(k = -\dfrac{c}{3}\)

Notes

B1ft: Deduces the correct inverse matrix. Follow through their \(c\) so allow for \(\dfrac{1}{3k + c}\mathbf{B}\) or if found determinant \(\dfrac{1}{\text{their det}}\mathbf{B}\)

Notes

M1: Complete method to find the values of \(x\), \(y\) and \(z\) using their inverse matrix

A1: At least one correct coordinate simplified or unsimplified.

A1: All coordinates correct and simplified. Condone as a column vector. Does not need to be written as a coordinate.

SC If candidate writes \(\dfrac{1}{3k + 1}\begin{pmatrix}4k - 2 & 1 & 7k - 4\\ -10 & 3 & -19\\ 3 - k & -1 & 6 - k\end{pmatrix}\begin{pmatrix}10\\ 3\\ 1\end{pmatrix}\) but ends up with at least one of \(x = 47k - 21,\ y = -110,\ z = 33 - 11k\) scores M1 A1 A0

Notes

B1: Correct value, check within the question. If more than one value is stated the correct value must clearly be selected.

Notes

M1: Uses their answer from part (a) to determine a value for \(k\) so that \(k \times\)their 210 is divisible by 360. If their answer to part (a) is in radians \(k \times \text{their } \dfrac{7\pi}{6}\) is divisible by \(2\pi\).

A1: Correct value must be using an angle of 210. There may be no working but allow M1 A1 for \(k = 12\) following an answer of 210 or \(\dfrac{7\pi}{6}\) in part (a)

Note: an angle of 30, 150, 330 also gives \(\boldsymbol{k = 12}\) but is M1A0

Notes

B1: Correct matrix

Notes

M1: Complete method to find the coordinates of \(B\). Look for at least two correct terms for each multiplication, follow through when multiplying by N.
Alternatively finds NM (not MN), look for 4 correct non zero terms if no method is shown and then multiplies to find the coordinates of \(B\).

A1*: Correct coordinates (condone vector) Must have been working with exact values throughout. If working in decimals M1 A0
It is insufficient to just write \(\mathbf{NM}\begin{pmatrix}-2\\ 4\\ 3\end{pmatrix} = \begin{pmatrix}\sqrt{3} + 2\\ 2\sqrt{3} - 1\\ 3\end{pmatrix}\) there must be some evidence of matrix multiplication seen.

Notes

M1: Identifies and applies an appropriate strategy to find the required angle e.g. cosine rule or scalar product. Note: \(AB^2 = 56 - 12\sqrt{3}\) and \(AB = 3\sqrt{6} - \sqrt{2} = 5.93\ldots\)

A1*: Correct value from correct equation

Notes

M1: Uses the given angle with \(\dfrac{1}{2}ab\sin C\) with appropriate \(a\), \(b\) and \(C\)
Outside spec: uses the cross product \(\dfrac{1}{2}\left|\overrightarrow{OA} \times \overrightarrow{OB}\right|\)

A1: Correct area to 3 significant figures

Notes

(Corrected from the printed mark scheme: the completed-square, quadratic-formula and vertex lines are printed with \(-2k^2 - 10k - 25\), \(-2(k + 2.5)^2 - 12.5\), \(-4(-2)(-25)\), "determinant = −5.75" and "maximum value of determinant = −5.25". The determinant is \(-2k^2 - 10k - 24 = -2(k + 2.5)^2 - 11.5\), so these are typed with −24, −11.5 and a maximum value of −11.5.)

M1: Correct method to find the determinant, condone a single sign slip but not on second term must be +2 (…)
Note: May expand along any row or column.
A1: Correct simplified determinant

M1: Either

• Finds the value of the discriminant or sufficient working seen to identify the sign e.g. 100 – 192
• Completes the square an rearranges so that \((k \pm a)^2 = -b\)
• Completes the square and states that \((k \pm a)^2 \geqslant 0\)
• Completes the square and states that \(-\alpha(k \pm a)^2 \leqslant 0\)
• Differentiates the determinant to find the coordinates of the vertex
• Use the quadratic formula to find the imaginary roots

A1: Correct solution only
Either

• Correct value for the discriminant (may be implied), concludes less than 0, therefore no real roots and non singular.
• Correct completing the square and conclude no real roots as square root of negative therefore non singular
• Correct completing the square and shows > 0 therefore no real roots and non singular.
• Correct completing the square and shows < 0 therefore no real roots and non singular.
• Correct coordinates of the vertex and negative quadratic therefore no real roots and non singular.
• Use the quadratics formula to find the correct imaginary roots therefore no real roots/value for \(k\) and non singular.

Note \(k = \dfrac{-5 \pm \sqrt{23}\mathrm{i}}{2}\) which is not real is M0 A0 unless uses the quadratic formula or completing the square to show where this has come from

Notes

M1: Uses matrix Q to find the coordinates of the points \(A'\) and \(B'\). Condone a sign slip.
A1: Correct coordinates for the points \(A'\) and \(B'\), they do not need to be labelled
M1: Finds the distance between their points \(A'\) and \(B'\) which must not be equal to \(A\) and \(B\), sets equal to \(\sqrt{58}\), forms a 3TQ.
A1: Correct 3TQ form correct coordinates
A1: Correct values cso

Misread: A common misread is 3 instead of – 3, the first 3 mark only can be scored using the misread rule

M1: \(\begin{pmatrix}2 & -1\\ 3 & 0\end{pmatrix}\begin{pmatrix}a & 4\\ 2 & -a\end{pmatrix} = \begin{pmatrix}\ldots & \ldots\\ \ldots & \ldots\end{pmatrix}\)

A1: \(\begin{pmatrix}2a - 2 & 8 + a\\ 3a & 12\end{pmatrix}\) or \((2a - 2,\ 3a)\) and \((8 + a,\ 12)\)

M1: \(\sqrt{\left[(2a - 2) - (8 + a)\right]^2 + \left[3a - 12\right]^2} = \sqrt{58}\)

A0, A0 This does lead to the correct answer but can score the first three marks only.

Notes

B1: Any valid limitation – see scheme for some examples. Must refer to a feature of the categories given.

Notes

(b)(i)
M1: Attempts to use the given information to set up a matrix equation and find the numbers of mammals after one month e.g.

or attempts to square the matrix to find the number of mammals after two months e.g.

dM1: Forms an equation, in their variable for number of breeders at the start, setting their number of newborns after 2 months equal to 48 and solves for their variable to find the initial number of breeders.

A1: For identifying 25 mammals at the start of the study. Allow 25 mammals or just 25 or e.g. \(B_0 = 25\) so ignore how they label it just look for 25

Note that in some cases work may be minimal e.g.

(ii)
A1*: For correctly showing \(a = 0.8\). Must see the correct work to establish the correct value or equivalent by verification with a minimal conclusion e.g.

Notes

B1: Deduces correct determinant for the matrix. Allow equivalents e.g. \(\dfrac{96}{125}\) May be implied.

M1: Recognisable attempt at the adjoint matrix. Look for at least 3 non-zero entries correct.

A1: Correct inverse. Accept awrt \(-2.6b\) for the upper right entry and awrt 2.08 for middle right entry, or accept with determinant still outside. Apply isw once a correct answer is seen.

Notes

(d) Alternative:

Notes

M1: Attempts (their inverse matrix) \(\times \begin{pmatrix}596\\ 464\\ 437\end{pmatrix}\) correctly and adds the 3 expressions together to find the total in terms of \(b\).

M1: Sets their total = 1015 and solves for \(b\).

A1: awrt 0.45

Alternative:
M1: Uses the original matrix with \(a = 0.8\) and \(\begin{pmatrix}596\\ 464\\ 437\end{pmatrix}\) to form 3 equations in their variables and \(b\) and uses these and the 1015 to find the number of Newborns.
M1: Uses their values in the \(y\) component and solves for \(b\).
A1: awrt 0.45

Notes

M1: Defines new variables for male and female newborns (accept if a clear notation is used if not defined) and sets up a 4×4 matrix with structure shown, or male and female rows swapped, with the correct 0 entries in at least 4 places.

A1ft: Fully correct matrix system shown, accepting anything (including 0) for the unknown spaces shown – but must have all the 0’s and upper right entries correct. Accept \(b\) or their value of \(b\) in place of 0.45

Notes

B1: Shows true for \(n = 1\).
Need to see \(n = 1\) substituted into rhs. The minimum for B1 would be \(\begin{pmatrix}1 & -2 \times 1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}\).
There is no need to state “True for \(n = 1\)”

M1: (Assumes for \(n = k\) and) multiplies original matrix by \(k\)th power matrix either way round. Note that the assumption statement is not needed for this mark (but see below) so just look for:

A1: Achieves a correct unsimplified matrix.

A1: Reaches correct form for the matrix with the \(k + 1\) factored out with no errors and the correct unsimplified matrix seen previously. Note that the result may be proved by equivalence (see below).

A1: Correct conclusion with assumption made (which may be implied in their conclusion if they say “if true for \(n = k\) then…”). This mark is dependent on all the previous marks apart from the B mark and is gained by conveying all the underlined points.
Allow this mark to score as long as all the underlined points are seen as narrative in their solution.
There must be the assumption statement somewhere or the “if…then…” idea in the conclusion. If awarded for the assumption statement condone e.g. true for \(n = k\) in the conclusion.

The conclusion must convey the “if true for \(n = k\) then true for \(n = k + 1\)” idea and not e.g. true for \(k\), true for \(k + 1\), true for 1 therefore…but see the previous note.
For the “true for all \(n\)” part condone e.g. “true for \(n\)”, “true for all integers after 1”, “true for \(\mathbb{Z}^+\)”,
But do not allow “true for all values”, “true for all real numbers”

Q4 Extra Notes:

1. For candidates who use \(n\) instead of \(k\) throughout withhold the final mark if the work is otherwise correct.

2. For equivalence proofs, this would be minimally acceptable:

\(\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2 \times 1\\ 0 & 1\end{pmatrix}\) B1

“Need to prove” oe e.g. “target is”, “\(n = k + 1\)”, “need” etc. \(\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}^{k+1} = \begin{pmatrix}1 & -2k - 2\\ 0 & 1\end{pmatrix}\)

Note that there must be a reference to \(k + 1\) as above or e.g. (Target =) \(\begin{pmatrix}1 & -2(k + 1)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2k - 2\\ 0 & 1\end{pmatrix}\)

\(\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}^{k+1} = \begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & -2k\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2k - 2\\ 0 & 1\end{pmatrix}\) M1A1A1

Hence result is true for \(n = k + 1\). As true for \(n = 1\) and have shown if true for \(n = k\) then it is true for \(n = k + 1\), so it is true for all \(n\). A1

Without the “Need to prove…” oe the response would score B1M1A1 and then A1 if the factorised form was shown or equivalence shown and then A1 for the correct conclusion.

3. Allow e.g. “correct” for “true” in the conclusion

Notes

B1: Identifies the single transformation as a rotation only

B1: Correct angle and axis. Ignore any reference to direction.
Note \(x\)-plane, \(zy\)-plane and \(x = 0\) are 2^nd B0
Any additional incorrect statements is 2^nd B0

Notes

B1: Explains that they should be multiplied the other way around

Notes

B1: Correct exact matrix

Note: \(5\sqrt{3} \times \dfrac{\sqrt{3}}{2}\) must be simplified to \(\dfrac{15}{2}\)

Condone \(\dfrac{2\sqrt{3}}{2}\) not simplified

Notes

Alternative

Notes

M1: Evaluates \(\mathbf{M}^2\) and uses in the equation given.

A1: Correct value of \(a\) deduced

A1*: Correct work to show \(k = -9\). If off diagonals are used no further justification is needed (they are “given” the result is true). If the bottom right entry is used there must be a valid reason for rejecting \(-2\) as a solution (ie checking the off diagonal).

Alternative: Using \(k = -9\)
M1: Evaluates \(\mathbf{M}^2\) and uses in the equation given.
A1: Correct value of \(a\) deduced
A1*: Draws the conclusion that \(k = -9\)

Notes

M1: Sets up the matrix equation for invariant lines and extracts the simultaneous equations from the matrix equation.

M1: Eliminates “\(X\)” to get a linear equation in “\(x\)”.

A1: Correct equation need not be simplified isw

M1: Solves their quadratic equation in \(m\) by any valid means including calculator

A1: Deduces \(y = \dfrac{3}{5}x\) is a fixed line (where \(c = 0\)). If the value for \(m\) here is wrong, allow this A for \(y = -2x\) if the general case for the final A is not scored.

A1: Deduces \(y = -2x + c\) is a fixed line where \(c\) can be any value. Must include all the lines.

Note: \(y = \dfrac{3}{5}x\) and \(y = -2x\) scores final A1 A0

Special Case 1 M1M1A0M1A1A0

M1: Sets up the matrix equation for invariant lines and extracts the simultaneous equations from the matrix equation.

M1: Eliminates “\(X\)” to get a linear equation in “\(x\)”. \(\Rightarrow 6x - 9mx = -2mx + 5m^2x\)

A0: Incorrect equation.

M1: Solves their quadratic equation in \(m\) by any valid means.
\(\Rightarrow 5m^2 + 7m - 6 = 0 \Rightarrow (m + 2)(5m - 3) \Rightarrow m = -2, \dfrac{3}{5}\)

A1: Deduces \(y = \dfrac{3}{5}x\) is a fixed line (where \(c = 0\)). If the value for \(m\) here is wrong, allow this A for \(y = -2x\) if the general case for the final A is not scored.

A0: Incorrect equation

Special Case 2 Finding the line of invariant points M1M0A0M0A1A0

M1: Sets up the matrix equation for a line of invariant points and extracts the simultaneous equations from the matrix equation.

M0 A0 M0

A1: Deduces \(y = \dfrac{3}{5}x\) is a fixed line

A0: Incorrect equation

Outside the specification

M1: Find the eigenvalues \(\begin{pmatrix}-2 - \lambda & 5\\ 6 & -9 - \lambda\end{pmatrix} \Rightarrow (-2 - \lambda)(-9 - \lambda) - 6 \times 5 = 0\) leading to a 3TQ and solves to find a value for \(\lambda\), \(\lambda^2 + 11\lambda - 12 = 0 \Rightarrow \lambda = \ldots\{1, -12\}\)

M1: Uses one of their eigenvalues to find an equation \(\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = -12\begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow \ldots\{y = -2x\}\) or \(\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow \ldots\{5y = 3x\}\)

(corrected from the printed mark scheme: the factor \((-2 - \lambda)\) is printed as \((2 - \lambda)\), and the bottom right entry of the matrix in the two eigenvector equations is printed as \(9\) instead of \(-9\))

A1: One correct equation

M1: Uses both of their eigenvalues to find an equation

A1: Deduces \(y = \dfrac{3}{5}x\)

A1: Deduces \(y = -2x + c\)

Notes

B1: Identifies \(y = \dfrac{3}{5}x\) is a line of fixed points with reason. Allow if \(c = 0\) is assumed. See scheme for possible reason.

Notes

B1: \(y = 3\)

B1ft: Follow through on the value of \(z\) which comes from their \(y\) divided by 3

M1: A complete method to find the value of \(x\). Uses \(z - 3y = k\) to find a value for \(k\) then finds a value for \(x\) using \(x - 3z = k\) and their values for \(z\) and \(k\). Condone a slip with the coefficients if the intention is clear but must have the correct letters.
Alternatively uses \(x - 3z = k = z - 3y\) with values for \(y\) and \(z\) to find a value for \(x\).

A1: \(x = -5\)
Correct answers only scores full marks.

Notes

B1: Shows the statement is true for \(n = 1\). Accept as minimum \(\begin{pmatrix}1 - 6 & 9\\ -4 & 1 + 6\end{pmatrix} = \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\)

M1: Makes the inductive assumption, assume true \(\boldsymbol{n = k}\). This may appear in the conclusion.

M1: A correct statement for \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^{k+1}\) in terms of \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^k\), can be either way round.
Can be implied by \(\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\times\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\) or \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\times\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\)

M1: Carries out the multiplication correctly, condone sign slips

A1: Correct simplified matrix from fully correct working

A1: Completes the inductive argument by showing clearly the matrix has the correct form (must have \((k + 1)\) factors in terms) or uses the result with \(n = k + 1\) and shows that their result is the same.
Conclusion conveying all three underlined points or equivalent at some point in their argument. Depends on all three M’s and A marks but can be scored without the B mark as long as it is stated true for \(n = 1\)

Notes

M1: Finds the determinant of the matrix \(\mathbf{M}\). Must be seen in part (a). Allow one slip if no method shown.

A1: Correct value for determinant, states doesn’t equal 0 (accept \(\gt 0\)) and draws the conclusion that the matrix is non-singular. If non-singular meaning determinant is non-zero is given in a preamble then accept a minimal conclusion (e.g. “hence shown”), but there must be a conclusion.

Notes

M1: Finds the matrix of minors, at least 5 correct values.

M1: Finds the matrix of cofactors and transposes (in either order). Note: some will do all these steps in one go, which is fine as long as it is clear what they have done. Allow minor slips if the process is clearly correct.

M1: Completes the process to find the inverse matrix, divides by the determinant.

A1: Correct matrix.

Notes

B1: Coordinates of \(Q\) correctly stated, accept as a column vector.

Notes

M1: Correct attempt to find coordinates of \(R\) using the given matrix with \(\theta = 120\). Must be multiplying in the correct way round. With no working two correct values or (−2.27, 3, −7.93) implies this mark.

A1: Correct exact coordinates as shown in scheme. Accept as a column vector. Cos 120 and sin 120 must have been evaluated.

Notes

M1: Applies the distance formula with the coordinates of \(P\) and their \(R\). Alternatively finds the vector \(\overrightarrow{PR}\) or \(\overrightarrow{RP}\) then applies length of a vector formula.

A1: Correct answer following correct coordinates of \(R\), must be a surd but need not be fully simplified.

Notes

B1ft: Shows the dot product is zero between the vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) and draws the conclusion perpendicular. Accept with \(\pm\) vectors for each. Follow through as long as the vectors are of the correct form, so \(\overrightarrow{PR} = \begin{pmatrix}a\\ 0\\ b\end{pmatrix}\) and \(\overrightarrow{PQ} = \begin{pmatrix}0\\ c\\ 0\end{pmatrix}\)

Note They could state if vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) are perpendicular then \(\overrightarrow{PR}.\overrightarrow{PQ} = 0\) then shows \(\overrightarrow{PR}.\overrightarrow{PQ} = 0\) this is B1

Notes

M1: Correct method for the area of the triangle, follow through on their coordinates of \(R\) and \(Q\). May see longer methods if they do not realise the triangle is right angled.

A1: For \(6\sqrt{51}\) cso following correct coordinates of \(R\)

Alternative 1

M1 Complete method to find the correct area
Finding all the lengths \(|PQ| = 6,\ |PR| = \sqrt{204} = 2\sqrt{51},\ |QR| = \sqrt{240} = 4\sqrt{15}\)
Use cosine rule to find an angle e.g. \(\cos PRQ = \dfrac{240 + 204 - 36}{2 \times \sqrt{240} \times \sqrt{204}} = \dfrac{\sqrt{85}}{10}\)
leading to \(PRQ = 22.7\ldots\) or \(\sin PRQ = \sqrt{1 - \left(\dfrac{\sqrt{85}}{10}\right)^2} = \ldots\left\{\dfrac{\sqrt{15}}{10}\right\}\)
Uses the area of the triangle \(= \dfrac{1}{2} \times \sqrt{240} \times \sqrt{204} \times \dfrac{\sqrt{15}}{10}\) or \(= \dfrac{1}{2} \times \sqrt{240} \times \sqrt{204} \times \sin 22.8\)

(Corrected from the printed mark scheme: the printed scheme has \(|PR| = \sqrt{240} = 4\sqrt{15}\) and \(|QR| = \sqrt{204} = 2\sqrt{51}\); these two lengths are the wrong way round, since part (c) gives \(|PR| = \sqrt{204}\).)

A1: For \(6\sqrt{51}\)

Alternative 2

M1: Uses \(\dfrac{1}{2}|\mathbf{a} \times \mathbf{b}|\) to find the required area
e.g. \(QP = \begin{pmatrix}0\\ 6\\ 0\end{pmatrix}\quad RP = \begin{pmatrix}12 - \sqrt{3}\\ 0\\ 3 + 4\sqrt{3}\end{pmatrix}\) cross product\[\begin{vmatrix}0 & 6 & 0\\ 12 - \sqrt{3} & 0 & 3 + 4\sqrt{3}\end{vmatrix} = 6\left(3 + 4\sqrt{3}\right)\mathbf{i} - 6\left(12 - \sqrt{3}\right)\mathbf{k}\]\[\text{Area} = \frac{1}{2}\sqrt{\left(6\left(3 + 4\sqrt{3}\right)\right)^2 + \left(-6\left(12 - \sqrt{3}\right)\right)^2} = \frac{1}{2}\sqrt{7344}\]

(Corrected from the printed mark scheme: the printed cross product is \(-6\left(12 - \sqrt{3}\right)\mathbf{i} + 6\left(3 + 4\sqrt{3}\right)\mathbf{k}\), with the \(\mathbf{i}\) and \(\mathbf{k}\) components the wrong way round. The area is unchanged.)

A1: For \(6\sqrt{51}\)

Notes

B1: Shows that the result holds for \(n = 1\). Must see substitution in the RHS minimum required is \(\begin{pmatrix}3 & \frac{a}{2}(3 - 1)\\ 0 & 1\end{pmatrix}\) and reaches \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}\)

M1: Assumes the result is true for some value of \(n = k\). Assume (true for) \(n = k\) is sufficient. Alternatively states assume \(\mathbf{M}^n\) or \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}^k = \begin{pmatrix}3^{k} & \dfrac{a}{2}\left(3^{k} - 1\right)\\ 0 & 1\end{pmatrix}\)

M1: Sets up a matrix multiplication of their assumed result multiplied by the original matrix, either way round. Allow a slip as long as the intention is clear.

A1: Achieves a correct un-simplified matrix

A1: Reaches a correct simplified matrix with no errors, the correct un-simplified matrix seen previously and at least one intermediate line which must be correct.

A1: Correct conclusion. This mark is dependent on all previous marks except B mark but \(n = 1\) must have been attempted. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Condone \(n \in \mathbb{Z}\)

Notes

(i) B1: States correct determinant. This can be implied by a correct equation

M1: Correct method to find a value of \(n\) using \(5 \times \text{`their } \det\left(\mathbf{M}^n\right)\text{'} = 1215\) which involves solving an index equation of the form \(p^n = q\) where \(n \gt 1\)

A1: \(n = 5\)

(ii) M1: Sets up an equation by multiplying the matrix \(\mathbf{M}^n\) by \(\begin{pmatrix}2\\ -2\end{pmatrix}\) setting equal to \(\begin{pmatrix}123\\ -2\end{pmatrix}\) and reaches a value for \(a\). You may just see \(2\left(3^n\right) - 2\dfrac{a}{2}\left(3^n - 1\right) = 123 \Rightarrow a = \ldots\)
Follow through on their value for \(n\).

A1: \(a = 1.5\)

(Corrected from the printed mark scheme: in the inverse-matrix line the printed scheme has \(a\) in place of 243 in the bottom right of the inverse matrix, and the equation as \(\dfrac{123 - 2\frac{a}{2}(243 - 1)}{243} = -2\).)

Notes

Mark (i) and (ii) together

(i) B1: Defines 3 variables, minimum e.g. construction = \(C\), Design = \(D\), Hospitality = \(H\). This may be seen in text of the question, abbreviations may be used

(ii) M1: Finds either the increase or the number of students in 2020. This may be implied by any equation which equals 1113 or 1112.997. If students use 1100 instead of 1110 this is slip and we can award this mark.

M1: Attempts to use the model to set up at least 2 equations

A1: All 3 simplified equations correct (decimals or fractions), one for each different piece of information. Award with mark even if B0 is scored and it is clear what the variables used stand for. Ignore any additional equations even if incorrect. As soon as 3 correct equations are seen you may award this mark.

Alternative approach
(i) B1: Construction = \(H + 370\), Design = \(D\), Hospitality = \(H\)
(ii) M1M1A1: \(H + 370 + D + H = 1110\) o.e \(\quad C = H + 370 \quad 1.0125(H + 370) + 1.025D + 0.98H = 1113\) or 1112.997 o.e. they do not need to be simplified

Special Case: Forming an equation in one variable
(a)(i) B1: Hospitality = \(x\), Construction = \(x + 370\), Design = \(740 - 2x\)
(ii) M1M1A1: \(1.0125(x + 370) + 1.025(740 - 2x) + 0.98x = 1113\) or 1112.997

(a)(i) B1: Hospitality = \(x - 370\), Construction = \(x\), Design = \(1480 - 2x\)
(ii) M1M1A1: \(1.0125(x) + 1.025(1480 - 2x) + 0.98(x - 370) = 1113\) or 1112.997

(b) M0A0M0A0: They have an equation and are not forming and solving a matrix equation

Notes

This is M1 M1 A1 A1 on ePen but is marked M1A1M1A1

M1: Uses their equation in part(a) to set up a matrix equation of the form \(\begin{pmatrix}\ldots & \ldots & \ldots\\ \ldots & \ldots & \ldots\\ \ldots & \ldots & \ldots\end{pmatrix}\begin{pmatrix}C\\ D\\ H\end{pmatrix} = \begin{pmatrix}\ldots\\ \ldots\\ \ldots\end{pmatrix}\), where “…” are numerical values.

A1ft: Correct matrix equation for their equations

dM1: Dependent on previous method mark. Writes \(\left(\text{their } \mathbf{A}\right)^{-1}\begin{pmatrix}1110\\ \text{their ``370''}\\ \text{their ``3''}\end{pmatrix}\) and obtains at least one value of \(C\), \(D\) or \(H\). The inverse matrix need not be found, writing \(\mathbf{A}^{-1}\begin{pmatrix}1110\\ 370\\ \text{their ``3''}\end{pmatrix} = \ldots\) is sufficient. A correct matrix equation followed by correct values implies this mark.
Condone \(\begin{pmatrix}1110\\ \text{their ``370''}\\ \text{their ``3''}\end{pmatrix}\mathbf{A}^{-1} = \ldots\) as long as they reach some values. The values imply the correct method

Note: \(\begin{pmatrix}1 & 1 & 1\\ 1 & 0 & -1\\ 0.0125 & 0.025 & -0.02\end{pmatrix}^{-1} = \begin{pmatrix}\frac{10}{23} = 0.43\ldots & \frac{18}{23} = 0.78\ldots & -\frac{400}{23} = -17.39\ldots\\[4pt] \frac{3}{23} = 0.13\ldots & -\frac{13}{23} = -0.56\ldots & \frac{800}{23} = 34.78\ldots\\[4pt] \frac{10}{23} = 0.43\ldots & -\frac{5}{23} = -0.21\ldots & -\frac{400}{23} = -17.39\ldots\end{pmatrix}\)

\(\begin{pmatrix}1 & 1 & 1\\ 1 & 0 & -1\\ 1.0125 & 1.025 & 0.98\end{pmatrix}^{-1} = \begin{pmatrix}\frac{410}{23} = 17.82\ldots & \frac{18}{23} = 0.78\ldots & -\frac{400}{23} = -17.39\ldots\\[4pt] -\frac{797}{23} = -34.65\ldots & -\frac{13}{23} = -0.56\ldots & \frac{800}{23} = 34.78\ldots\\[4pt] \frac{410}{23} = 17.82\ldots & -\frac{5}{23} = -0.21\ldots & -\frac{400}{23} = -17.39\ldots\end{pmatrix}\)

A1: Interprets the answer in the context of the question, minimum is \(C = 720\), \(D = 40\), \(H = 350\) with their variables. Condone the variable not been defined for this mark if it is clear which variable belong to what course.

Note: they must be using a matrix equation to solve the equation to score any marks.

Alternative approach
For example
Equations simplifies to \(C - H = 370\), \(D + 2H = 740\) and \(1.025D + 1.9925H = 738.375\)
which leads to \(\begin{pmatrix}0 & 1 & 2\\ 1 & 0 & -1\\ 0 & 1.025 & 1.9925\end{pmatrix}\begin{pmatrix}C\\ D\\ H\end{pmatrix} = \begin{pmatrix}740\\ 370\\ 738.375\end{pmatrix}\) then \(\begin{pmatrix}C\\ D\\ H\end{pmatrix} =\)
\(\begin{pmatrix}17.826 & 1 & -17.3913\\ -34.6521 & 0 & 34.7826\\ 17.826 & 0 & -17.3913\end{pmatrix}\begin{pmatrix}740\\ 370\\ 738.375\end{pmatrix} = \begin{pmatrix}720\\ 40\\ 350\end{pmatrix}\)

Note: A 2 x 2 matrix is fine if it is appropriate for their equation.

Notes

(a)(i)
M1: Attempts to find AB. Usually this will be done on calculator so answer implies the method. If answer is incorrect allow for at least 6 correct entries or calculations shown.
This mark can be implied by a correct matrix for \(\mathbf{AB} - 3\mathbf{C}\) gives the first M1

M1: Uses their AB and 3C matrices to find a multiple I and states a value for \(k\)

A1: Correct proof with \(k = -24\) seen explicitly (may be in equation).

Minimum working required is \(\mathbf{AB} - 3\mathbf{C} = \begin{pmatrix}24 & 0 & 0\\ 0 & 24 & 0\\ 0 & 0 & 24\end{pmatrix}\) gets M1 then states a value for \(k\) M1 then \(k = -24\) gets A1

Special case: If minimum working required is not seen and just \(k = -24\) stated then M1 M0 A0 as they have not shown that the value of \(k\) works.

(ii)
B1: Correct explanation referring to the dimensions of BA and C (or 3C) and that they do not match in the equation. They can find both these matrices and then comment they cannot be subtracted.

Notes

(b) Mark (i) and (ii) altogether

M1: States or implies use of the correct method of using the inverse matrix.

M1: Carries out the process of multiplying after finding the inverse. May find inverse long hand first. Finding the inverse matrix then writes down an answer gains M1.

Note: There is no need to find the inverse matrix. If the inverse matrix is not stated just answers written down then two out of the three correct ordinates imply the M1.

A1: Correct solution. Must be clear that \(x = \dfrac{7}{2},\ y = 3,\ z = -\dfrac{5}{2}\) allow \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}3.5\\ 3\\ -2.5\end{pmatrix}\)

Note: If they solve using simultaneous equations only this is M0 M0 A0
If there is no reference to the inverse matrix and correct answers stated this is M0 M0 A0

Notes

(a) B1: Comments that the number of columns of matrix A (2) equals the number of rows of matrix B (2) therefore it is possible. Accept other terminology that is clear in intent e.g. “length of A” and “height of B”

(b) B1: Comments that matrix A and matrix B have different dimensions therefore it is not possible.

Notes

(a) B1: Deduces the correct value for \(\lambda = 5\)

B1: Deduces the correct values for \(a\) and \(b\)

(b) B1ft: Identifies and applies a correct method find the inverse matrix. May multiply from the given equation, in which case follow through on their value of lambda. Alternatively, award for a correct matrix found by calculator or long hand having found \(a\) and \(b\) and using these values in the matrix.

Notes

M1: A complete method to find the determinant of the matrix and sets it equal to 0

A1: Correct equation

M1: Uses appropriate correct trig identities to solve the equation and finds a value for \(\theta\)

A1: All three correct values \(\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}\) and no others in the range.

Notes

M1: Sets up a matrix equation in an attempt to find a fixed line and extract at least one equation.
A1: Correct equations.
M1: Eliminates \(X\) from the simultaneous equations and equates the coefficients of \(x\) leading to a quadratic equation in terms of \(m\).
dM1: Dependent on the previous method, finds the value of the discriminant, this can be seen in an attempt to solve the quadratic using the formula. Alternatively solves \(3c = -2mc + c\) and finds a value for \(m\).
Note: If the quadratic equation in \(m\) is solved on a calculator and complex roots given this is M0 as they are not showing why there are no real roots.
A1: Correct expression for the discriminant, states \(\lt 0\) and draws the required conclusion. Alternatively, correct value for \(m\), shows a contradiction in \(5 + 3m = 4m - 2m^2\) and draws the required conclusion.

Alternative
M1: Sets up a matrix equation in an attempt to find a fixed line and extract at least one equation.
A1: Correct equations.
M1: Eliminates \(X\) from the simultaneous equations and equates the coefficients of \(x\) leading to a quadratic equation in terms of \(m\).
dM1: Dependent on the previous method, finds the value of the discriminant.
A1: Correct expression for the discriminant, states \(\lt 0\) and draws the required conclusion.

Notes

Way 1

M1: A full method to find \(k\) such as attempting the square root of the determinant of \(\mathbf{M}\). It is immediately deducible so the method may be implied by \(k = 8\).

A1: \(k = 8\)

M1: A full method to find a value of \(\theta\) using their \(k\), no need to justify quadrant. Only one equation needed for this mark. Allow if a radians answer is given. May be implied by a correct angle.

A1: Correct angle in degrees.

Way 2

M1: Multiplies the correct matrix representing transformation \(Q\) by the matrix representing transformation \(P\) and sets equal to matrix \(\mathbf{M}\). Allow for the matrices either way round as the transformations commute. No need to see the identity matrix, just multiplying through by \(k\) is sufficient.

A1: Both correct equations. Note that if a correct value of \(k\) is found, this A is scored under Way 1.

M1: Solves their simultaneous equations to find a value for \(\theta\) (or \(k\))

A1: \(\theta = 120^\circ\) and \(k = 8\)

Notes

M1: Complete method to find the area of \(S^{\prime}\): \(\text{``}\text{their } k^2\text{''} \times \text{``}\text{their } 2a^2\text{''}\). Must be an attempt at the area of \(S\) but it need not be correct.

A1ft: Deduces the correct area for \(S^{\prime}\), follow through their value of \(k\)

Notes

(Corrected from the printed mark scheme: the printed example is \(2(1 - 4)b + (1 - 4) = 0\); from \(2(a - 4) + b(a - 4) = 0\) with \(a = 1\) it is \(2(1 - 4) + b(1 - 4) = 0\).)

M1: Forming two equations, one involving \(a\) only and one involving \(a\) and \(b\)

dM1: Dependent on previous mark, solves a 3TQ involving \(a\)

A1: Correct values for \(a\)

dM1: Dependent on first method mark Substitutes one of their values of \(a\) into an equation involving \(a\) and \(b\) and solve to find a value for \(b\). Alternatively factorises either \(2a + ab = 0\) and uses \(a \neq 0\) to find a value for \(b\).

A1: Correct value for \(b\)

Alternative (i)(a)

M1: Finds \(\mathbf{A}^{-1}\) and sets equal to \(\mathbf{A}\) and forms two different equations

dM1: Dependent on previous mark. Eliminates \(b\) from their equations and solves a 3TQ involving only the constant \(a\). Alternatively if the value of \(b\) is found first substitutes their value for \(b\) into their determinant = –1 to form and solve a 3TQ for \(a\)

A1: Correct value for \(a\)

dM1: Dependent on first method mark. Substitutes a value for \(a\) into an equation to find a value for \(b\). Alternatively uses one equation to find the determinant = –1 and uses this to find a value of \(b\).

A1: Correct values for \(b\)

Notes

M1: Extracts simultaneous equations using their matrix A with their smaller value of \(a\).

M1: Gathers terms from their two equations.

A1: Achieves the correct equations and deduces the correct line. Accept equivalent equations as long as both have been shown to be the same.

Notes

B1: Area of the triangle \(T = 3\)

M1: Full method. Finds the determinant, sets equal to 15/their area and solves the resulting 3TQ

A1: Correct quadratic

A1: \(p = 1\) only

Notes

B1 One correct row or column

B1: All correct

Notes

M1: Multiplies the matrices QP in the correct order (if answer only then evidence can be taken from 3 correct or 3 correct ft elements)

A1ft: Correct matrix following through on their answer to part (b) and their value of \(p\) as long as it is a positive constant

Notes

(a)
M1: Correct method to find the determinant. Condone one sign slip
M1: A correct first step in obtaining the inverse. Could be the matrix of minors or cofactors. Condone sign slips as long as the intention is clear.
M1: Fully correct method to obtain the inverse. Attempts matrix of minors, cofactors, transposes and 1/determinant
A1: Correct matrix

Notes

(b)
M1: A complete strategy for solving the given equations e.g. multiplies the given coordinates by their inverse or solves simultaneously to achieve values for \(x\), \(y\) and \(z\)
A1ft: Correct calculation on their inverse matrix (unsimplified) or at least on correct value if solving simultaneously
A1: Correct coordinates

Notes

(c)(i)
M1: Uses a correct strategy that will lead to establishing a value for \(q\). E.g. eliminating one of \(x\), \(y\) or \(z\)
M1: Solves a suitable equation to obtain a value for \(q\)
A1: Correct value

(c)(i) Alternative 1
M1: Equating coefficients leading to two out of three equations and solves to find values for a and b
M1: Solves a suitable equation to obtain a value for \(q\) using their values for \(a\) and \(b\)

(c)(i) Alternative 2
M1: Finds a coordinate of intersection of the planes \(4x + 5y + 7z = 1\) and \(2x + y - z = 2\)
M1: Substitutes the values for \(x\), \(y\) and \(z\) into \(x + y + z = q\) to reach a value for \(q\)
A1: Correct value

(ii)
M1: Uses a correct strategy to obtain the Cartesian equation of the line or the general coordinates
A1: Correct Cartesian equation or coordinates in terms of a parameter.
M1: Uses their Cartesian equation to correctly extract the position and direction to form a vector equation for the required line
A1: Correct equation (o.e.) look out for multiples of the direction vector

Alternative (ii)
M1: Finds two different coordinates that lie on the line of intersection
A1: Correct coordinates
M1: Uses their coordinates to find the vector equation of the line that passes through them.
A1: Correct equation (o.e.), look out for multiples of the direction vector. Must have \(\mathbf{r} = \ldots\)

Alternative (ii) outside spec
M1: Finds the cross product between the normal vectors of two of the planes and a coordinate that lies on all three planes. If a coordinate is found in (i) it must be used in this part to award this mark.
A1: Correct cross product
M1: Uses the coordinate and the cross product to find the equation of the line
A1: Correct equation (o.e.), look out for multiples of the direction vector. Must have \(\mathbf{r} = \ldots\)

(corrected from the printed mark scheme: in (c)(ii), \(\lambda + 3z = -1\) is printed as \(\lambda + 3z = 1\), and \(x = \dfrac{5}{6} - \dfrac{2}{3}\lambda\) is printed as \(x = \dfrac{5}{5} - \dfrac{2}{3}\lambda\))

(Note: the cross products in the “outside the spec” alternative are as printed. The correct values are \((4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \times (\mathbf{i} + \mathbf{j} + \mathbf{k}) = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), \((4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - \mathbf{k}) = -12\mathbf{i} + 18\mathbf{j} - 6\mathbf{k}\) and \((\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - \mathbf{k}) = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), each a multiple of \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\).)

Notes

M1: Finds the determinant of the matrix corresponding to the system of equations.

A1: Sets determinant = 0 and solves their 3TQ to achieve \(k = 2\) (\(k = -8\) must be rejected)

Special case

M1A0: Uses \(k = 2\) and finds the determinant of the matrix corresponding to the system of equations
Shows that determinant = 0 and concludes that when \(k = 2\) there is no unique solution

Notes

M1: A complete method eliminating one variable from the equations using two different pairs of equations. Condone if a different value of \(k\) is used

A1: Achieves two equations in the same two variables

A1: Must give a reason, shows that the equations are a linear multiple of each other therefore they are consistent.

Alternative

(Note: the substitution shown uses \(5x + 2y = 1\) and \(11x + 2z = -3\).)

M1: A complete method eliminating one variable from the equations using two different pairs of equations. Substitutes these equations into one of the original equations in three variables.

A1: Achieves two correct equations in two different variables

A1: Shows that the equation works therefore they are consistent.

Notes

B1: The three planes form a sheaf. They must have full marks in (b) to award this mark.

Notes

B1: Correct interpretation. Need not mention survival but must be clear it is the (proportion of) juveniles that remain as juveniles the next year (ie those that survive but don’t progress to adulthood). E.g. accept “(number of) juveniles who do not become adults” but do not accept “surviving juveniles”.

Notes

Mark part (b) as a whole.

M1: Attempts the determinant in terms of \(a\). Allow miscopies for the attempt. Allow \(0.82a - 0.12\) as a slip.

M1: Attempts the form of the inverse, swapped leading diagonals and sign changed on both off diagonals. Allow miscopies of the numbers but the signs must be correct.

A1: Correct inverse matrix

Notes

M1: Use the inverse matrix and attempts to find the initial juvenile and adult populations. (May have determinant 1 for this mark.)
Alternatively, sets up simultaneous equations from the original system, \(15360 = aJ_0 + 0.15 \times A_0\) and \(43008 = 0.08 \times J_0 + 0.82 \times A_0\). Accept with \(J_n\) and \(A_n\) or other appropriate variables.

M1: Uses the sum of initial populations equals 64000 in an attempt to find \(a\). (May have determinant 1 for this mark.)
If using alternative, use of e.g. \(A_0 = 64000 - J_0\) in second equation to find \(J_0\), followed by attempt to find \(a\). Award for an attempt to solve the equations, but don’t be too concerned with the algebraic process as long as they are attempting to use all three equations.

A1: Correct value, \(a = 0.6\) (or 0.60 or \(\frac{3}{5}\)).

Notes

M1: Uses their \(a\) to find the value of \(J_0\). This mark may be gained for work done in (ii) if the alternative has been used but must have come from a correct method.

A1: Correct difference found, as long as there is no contradictory statement – so “decrease of 2560” is A0.

Notes

B1ft: Comments that the change is an increase so does not fit the model. Follow through their answer to (b) as long as at least a value for \(J_0\) has been found. If a decrease has been found allow for commenting the model is suitable. If an answer is given to (b)(iii), follow through on whatever their answer is. If no answer has been given, but an initial population found, a comparison should be made between this value and 15360 with conclusion must be consistent with their answer for \(J_0\)

(Corrected from the printed mark scheme: the value to compare with is printed as 153600; it is 15360.)

Notes

M1: Introduces a third category (may be Mature, Elderly or any suitable letter used) and sets up a matrix multiplication (the left hand side may be missing for this mark) with all three categories in the column vector. The dimension of the matrix should be 3 in at least either row or column, and there should be a \(3 \times 1\) vector.

A1: Sets up the new matrix equation, including both sides and making clear the zero (underlined) so that the correct progression that no new juveniles arise from the mature chimpanzees is clear. Overlook other values, though ideally the other two zeroes are shown too, to indicate mature chimpanzees do not regress to adulthood, and juveniles cannot proceed directly to mature chimpanzees.

Notes

M1: Attempts determinant, equates to zero and attempts to solve for \(k\) in order to establish the restriction for \(k\). For the determinant, at least 2 of the 3 “elements” should be correct.

May see rule of Sarrus used for determinant e.g.

\(|\mathbf{M}| = (2)(k)(-1) + (4)(3)(-1) + (3)(2)(1) - (3)(k)(1) - (2)(4)(2) - (-1)(3)(-1) = 0 \Rightarrow k = \ldots\)

(Corrected from the printed mark scheme: the fourth product is printed as \((3)(k)(-1)\); it should be \((3)(k)(1)\).)

A1: Describes the correct condition for \(k\) with no contradictions. Allow e.g. \(k \lt -5,\ k \gt -5\)

Notes

Way 1

M1: A complete strategy for solving the given equations. Need to see an attempt at the inverse followed by a correct method for finding \(x\), \(y\) and \(z\)

B1: Correct inverse matrix

M1: Uses their inverse and attempts the multiplication with the correct vector

A1: Correct values for \(x\), \(y\) and \(z\) in any form

A1ft: Correct values given in coordinate form only. Follow through their \(x\), \(y\) and \(z\).

Alternative: Way 2

M1: A complete strategy for solving the given equations. Need to see an attempt at eliminating one variable followed by a correct method for finding \(x\), \(y\) and \(z\)

B1: One correct value

M1: Uses the equations to find values for the other 2 variables

A1: Correct values for \(x\), \(y\) and \(z\) in any form

A1ft: Correct values given in coordinate form only. Follow through their \(x\), \(y\) and \(z\).

Notes

(c)(i)

M1: Uses a correct strategy that will lead to establishing a value for \(q\). E.g. eliminating one of \(x\), \(y\) or \(z\)

M1: Solves a suitable equation to obtain a value for \(q\)

A1: Correct value

(ii)

B1: Describes the correct geometrical configuration.

Must include the two ideas of planes and meeting in a line or forming a sheaf with no contradictory statements.

Notes

M1: An attempt to find \(\det(\mathbf{M})\). Just the calculation is sufficient. Sight of \(-18\) implies this mark, which may be embedded in an attempt at the inverse.

A1: \(\det(\mathbf{M}) = -18\) and reference to zero, e.g. \(-18 \ne 0\) and conclusion. The conclusion may precede finding the determinant (e.g. “Non-singular if \(\det(\mathbf{M}) \ne 0\), \(\det(\mathbf{M}) = -18 \ne 0\)” is sufficient or accept “Non-singular if \(\det(\mathbf{M}) \ne 0\), \(\det(\mathbf{M}) = -18\), therefore non-singular” or some other indication of conclusion.) Need not mention “\(\det(\mathbf{M})\)” to gain both marks here, a correct calculation, statement \(-18 \ne 0\), and conclusion hence \(\mathbf{M}\) is non-singular can gain M1A1.

Notes

M1: Recalls determinant is needed for area scale factor by dividing 63 by \(\pm\)their determinant.

A1ft: \(\dfrac{7}{2}\) or follow through \(\dfrac{63}{|\text{their det}|}\). Must be positive and should be simplified to single fraction or exact decimal. (Allow if made positive following division by a negative determinant.)

Notes

M1: Attempts the matrix multiplication shown or with equivalent, e.g. \(\begin{pmatrix}\frac{1}{2}y\\ y\end{pmatrix}\). May use \(\begin{pmatrix}x\\ y\end{pmatrix}\) and substitute \(y = 2x\) later and this is fine for the method.

A1: Correct multiplication and working leading to conclusion that the line is invariant. If the \(-6\) is not extracted, they must make reference to image points being on line \(y = 2x\). If the \(-6\) is extracted to show it is a multiple of \(\begin{pmatrix}x\\ 2x\end{pmatrix}\) followed by a conclusion “invariant” as minimum.

Alternative (Alt)

Marks as per main scheme.

Alternative (Alt 2)

M1: Identifies a point on the line \(y = 2x\) and finds its image under \(T\). If \((0, 0)\) is used there must be a clear statement it is because this is on the line, but for other points accept with any line on \(y = 2x\) without statement.

A1: Shows the image and another point, which may be \((0, 0)\), on \(y = 2x\) both map to points on \(y = 2x\), concludes line is invariant. Need not reference transformation being linear for either mark here.

Alternative (Alt 3)

M1: Attempts to find the equation of a general invariant line, or general invariant line through the origin (so may have \(c = 0\) throughout). To gain the method mark they must progress from finding the simultaneous equations to forming a quadratic in \(m\) and solving to a value of \(m\).

A1: Correct quadratic in \(m\) found, with \(m = 2\) as solution (ignore the other) and deduction that hence \(y = 2x\) is an invariant line. Ignore errors in the \((5m - 8)\) here as \(c = 0\) is always a possible solution. No need to see \(c = 0\) derived.

Notes

B1: Shows that the result holds for \(n = 1\). Must see substitution into the rhs.
The minimum would be: \(\begin{pmatrix}4 + 1 & -8\\ 2 & 1 - 4\end{pmatrix} = \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\).

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Sets up a correct multiplication statement either way round

A1: Achieves a correct un-simplified matrix

A1: Reaches a correct simplified matrix with no errors and the correct un-simplified matrix seen previously. Note that the simplified result may be proved by equivalence.

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Notes

(ii) Way 1

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1) – f(\(k\)) or equivalent work

A1: Achieves a correct expression for f(\(k\) + 1) – f(\(k\)) in terms of f(\(k\))

A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 2: \(\mathrm{f}(k + 1)\)

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1)

A1: Correctly obtains 4f(\(k\)) or \(21 \times 5^{2k-1}\)

A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 3: \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\)

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1) – \(m\)f(\(k\))

A1: Achieves a correct expression for f(\(k\) + 1) – \(m\)f(\(k\)) in terms of f(\(k\))

A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 4: \(\mathrm{f}(k) = 21M\)

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1)

A1: Correctly obtains \(84M\) or \(21 \times 5^{2k-1}\)

A1: Reaches a correct expression for f(\(k\) + 1) in terms of \(M\) and \(5^{2k-1}\)

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Notes

B1: Identifies the transformation as a rotation

B1: Correct angle. Allow equivalents in degrees or radians.

B1: Identifies the origin as the centre of rotation

These marks can only be awarded as the elements of a single transformation

Notes

B1: Shows the correct matrix in the correct form

Notes

M1: Multiplies the matrices in the correct order (evidence of multiplication can be taken from 3 correct or 3 correct ft elements)

A1ft: Correct matrix (follow through their matrix from part (b))

A correct matrix or a correct follow through matrix implies both marks.

Notes

M1: Translates the problem into a matrix multiplication to obtain at least one equation in \(k\) or in \(x\) and \(y\)

A1ft: Obtains one correct equation (follow through their matrix from part (c))

A1: Correct value for \(k\) in any form

B1: Checks their answer by independently solving both equations correctly to obtain \(2 + \sqrt{3}\) both times or substitutes \(2 + \sqrt{3}\) into the other equation to confirm its validity

Notes

B1: Evidence that the determinant is \(\pm 69\) (may be implied by their matrix e.g. where entries are not in exact form: \(\pm\begin{pmatrix}0.014 & 0.188 & 0.072\\ -0.159 & -0.072 & 0.203\\ -0.377 & 0.101 & 0.116\end{pmatrix}\)) (Should be mostly correct)

Must be seen in part (a).

B1: Fully correct inverse with all elements in exact form

Notes

M1: Any complete method to find the values of \(x\), \(y\) and \(z\) (Must be using their inverse if using the method in the main scheme)

A1: Correct coordinates

A solution not using the inverse requires a complete method to find values for \(x\), \(y\) and \(z\) for the method mark.

Correct coordinates only scores both marks.

Notes

B1: Describes the correct geometrical configuration.

Must include the two ideas of planes and meet in a point with no contradictory statements.

This is dependent on having obtained a unique point in part (b)

Typical solution

Typical solution

Let \(x = 0\):

Let \(y = 0\):

Two points on the line:

Typical solution

(i)

(ii)

so \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\)

Typical solution

When \(n = 1\):

so, the rule is true for \(n = 1\)

Assume the rule is true for \(n = k\)

so, the rule is also true for \(n = k + 1\)

So, by induction,

is true for all \(n \in \mathbf{N}\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

As \(\lambda = 1\) for a line of invariant points.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Anticlockwise rotation of \(30^\circ\) about the origin

Typical solution

LCM of 150 and 360

Typical solution

Typical solution

Typical solution

Typical solution

(i)

\((2, 5, -2)\)

(ii)

Shortest distance between \(l\) and \(B\) is

(iii)

Typical solution

Typical solution

30 cm²

Typical solution

Cofactors are

Typical solution

Typical solution

which is independent of \(k\)

Typical solution

Image of \(A\) is \((17, -38)\)

Typical solution

So \((0, 0)\) is the only invariant point

Typical solution

The new line is \(y = 2x + 8\)

Typical solution

T is a reflection in the line \(y = x\tan\dfrac{\pi}{3}\)

(3 d.p.)

Typical solution

Typical solution

Vector connecting \(L_1\) to \(L_2\)

Vector perpendicular to both lines

Alternative typical solution 1

Equation of \(L_2\): \(\mathbf{r} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix} + \mu\begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix}\)

General vector

Scalar product

Alternative typical solution 2

Parallel planes

Typical solution

Typical solution

For an invariant line \(y = kx\)

The equation has no real roots, so there is no invariant line of the form \(y = kx\).

Typical solution

\(\mathbf{A}\) and \(\mathbf{B}\) are commutative, hence \(\mathbf{AB} = \mathbf{BA}\)

Thus

Typical solution

Typical solution

So every image point lies in the plane \(x + 5y + 3z = 0\) when \(c = 3\)

Typical solution

(i)

(ii)

(iii)