If answer line can only be awarded B1 or B0 check the working space to see if more marks can be awarded
No misreads
B1 awarded for products must have matrices in correct order but can be recovered for B2 or B3 eg \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ -1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) is B0 but if recovered to \(\begin{pmatrix} 0 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ 0 \end{pmatrix}\) award B2
\(\begin{pmatrix} 0 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ 0 \end{pmatrix}\) may be seen as a single matrix eg \(\begin{pmatrix} 0 & -1 & -1 \\ -1 & -1 & 0 \end{pmatrix}\) or \(\begin{pmatrix} -1 & -1 & 0 \\ -1 & 0 & -1 \end{pmatrix}\)
\[\begin{pmatrix} 5 & 0 \\ 1 & -6 \end{pmatrix}\begin{pmatrix} 12 \\ 7 \end{pmatrix} = 15\begin{pmatrix} a \\ b \end{pmatrix}\]
Work out the values of \(a\) and \(b\). [2 marks]
Mark scheme
Answer
Mark
Comments
\((a =)\ 4\) and \((b =)\ -2\)
B2
B1 \((a =)\ 4\) or \((b =)\ -2\)
Additional guidance
Condone \(\begin{pmatrix} 4 \\ -2 \end{pmatrix}\) or \(15\begin{pmatrix} 4 \\ -2 \end{pmatrix}\) in working if nothing on answer line
B2
Condone \(\begin{pmatrix} 4 \\ k \end{pmatrix}\) or \(15\begin{pmatrix} 4 \\ k \end{pmatrix}\) or \(\begin{pmatrix} m \\ -2 \end{pmatrix}\) or \(15\begin{pmatrix} m \\ -2 \end{pmatrix}\) in working if nothing on answer line where \(k\) and \(m\) can be any number
\(-2c + 7\) and \(-10d + 21\) or \(-2c + 7 = -5\) or \(-10d + 21 = 0\)
M1
oe the two expressions may be seen in a matrix or embedded in two equations
\(c = 6\)
A1
\(d = \dfrac{21}{10}\) or \(d = 2.1\)
A1
oe value
Additional guidance
Ignore simplification or conversion attempt after correct answer seen for \(d\) eg \(d = \dfrac{-21}{-10}\) in working with \(d = -2.1\) on answer line
2nd A1
The two expressions may be seen in an incorrect matrix or two incorrect equations eg1 \(\begin{pmatrix} -2c + 7 & -10d + 21 \\ 2 & 9 \end{pmatrix}\)
M1
eg2 \(-2c + 7 = 3\) and \(-10d + 21 = 9\)
M1
A correct equation is not implied by a matrix equation eg \(\begin{pmatrix} -2c + 7 & 3 \\ 10d + 10 & 9 \end{pmatrix} = \begin{pmatrix} -5 & 3 \\ 0 & 9 \end{pmatrix}\)
M0
Mark scheme (b)
Answer
Mark
Comments
\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
B1
do not accept I or identity unless the correct 2 by 2 matrix is also seen
allow M1 for any three correct calculations or any three correct terms in the matrix calculations don’t need to be in a matrix condone missing brackets
may be written as \(k\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) where \(k = -2\)
Additional guidance
Errors in identity matrix will be penalised eg \(2\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\) or \(\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}\) are insufficient for A1
Must be evidence of identity matrix for A mark eg \(\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}\) and \(k = -2\)
Condone matrix written without brackets for M marks
M2A0
Correct multiplication of a unit square by both matrices will imply both M marks but won’t get the A mark unless the reflection matrix is shown. The matrix multiplications would need to be done in the correct order however
M2A0
Trying to operate matrices on a single point will only gain the first M mark (as it wouldn’t necessarily be true for all points). It would still require a correct matrix though
M1A0
Some candidates may multiply matrices in a grid. If \(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) is the result then the M1dep mark can be awarded. If not, do not award M1dep
Condone commas in matrices for M marks but will lose the A mark
(b) \(\begin{pmatrix} 7 & a^2 \\ b & -5 \end{pmatrix}\begin{pmatrix} 2 \\ a \end{pmatrix} = \begin{pmatrix} 78 \\ 12 \end{pmatrix}\)
Work out the values of \(a\) and \(b\). [3 marks]
Mark scheme (a)
Answer
Mark
Comments
Alternative method 1 Starts by multiplying 1st matrix by 3
\(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\)
B1
brackets may be missing but values must be in correct position in a 2 by 2 array
At least two values correct from evaluation of their \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\) \(\times\) \(\begin{pmatrix} 2 & 0 \\ -1 & 5 \end{pmatrix}\)
M1
brackets may be missing but values must be in correct position in a 2 by 2 array multiplication of matrices must be in the order shown
must have brackets ft \(3 \times\) their \(\begin{pmatrix} 6 & 10 \\ 2 & 0 \end{pmatrix}\) their \(\begin{pmatrix} 6 & 10 \\ 2 & 0 \end{pmatrix}\) must be a 2 by 2 array
For the final mark allow if there is intention to enclose the correct elements in brackets
Responses that start by multiplying 2nd matrix by 3 should be marked using the principles of Alt 1
Multiplying both matrices by 3 can score a maximum of B1 \(\begin{pmatrix} 12 & 6 \\ 3 & 0 \end{pmatrix}\) or \(\begin{pmatrix} 6 & 0 \\ -3 & 15 \end{pmatrix}\)
B1M0A0ft
Mark scheme (b)
Answer
Mark
Comments
\(14 + a^3 = 78\) or \(2b - 5a = 12\) or \(14 + a^3\) and \(2b - 5a\)
M1
oe eg \(a^3 = 64\) or \(2b + -5a = 12\) allow eg \(7 \times 2 + a^2 \times a\) for \(14 + a^3\) allow eg \(2 \times b - 5 \times a\) for \(2b - 5a\)
accept an exact value or a value rounded to 1 dp or better
Additional guidance
\(\begin{pmatrix} 14 + a^3 \\ 2b - 5a \end{pmatrix}\) or \((14 + a^3, 2b - 5a)\) with or without brackets
M1
\(a = 4\) (M1 is implied)
M1A1
M1 for \(2b - 5a = 12\) is implied by an incorrect value for \(a\) with a correct ft answer for \(b\) eg \(a = 8 \qquad b = 26\)
M1A0A1ft
An incorrect but exact value for \(a\) seen in working (eg \(\dfrac{8}{3}\)) with a rounded value for \(a\) on answer line (eg 2.6) Allow ft for \(b\) from the exact or the rounded value
\(a = 4\) and \(-4\) with one or both of \(b = 16\) and \(-4\)
M1A0A1ft
\(a = 4\) and \(-4\) (no values for \(b\) or incorrect values for \(b\))
Describe geometrically the single transformation represented by \(\mathbf{B}^2\) [2 marks]
Mark scheme (a)
Answer
Mark
Comments
Reflection in the \(x\)-axis or reflection in \(y = 0\)
B1
Additional guidance
Reflect(ed) in the \(x\)-axis
B1
Do not allow if there is additional incorrect information eg1 Reflection in the \(x\)-axis about the origin eg2 Reflection in the \(x\)-axis and rotated
B0 B0
Reflection
B0
Mark scheme (b)
Answer
Mark
Comments
Rotation through 180° centre the origin or enlargement scale factor \(-1\) centre the origin
B2
B1 \(\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\) or enlargement scale factor \(-1\) or rotation through 180° or indication that B represents rotation through 270° (anticlockwise centre the origin) or indication that B represents rotation through 90° clockwise (centre the origin)
Additional guidance
For B2 ignore any reference to clockwise or anticlockwise rotation
Condone omission of degrees symbol throughout eg B is rotation through 270
B1
Mark intention eg1 Rotate(d) 180 about \(O\) eg2 Enlarge(d) sf \(-1\)
B2 B1
Allow rotation through 540 centre the origin
B2
Do not allow if there is additional information that is incorrect eg1 Rotation through 180° and a reflection eg2 Enlargement sf \(-1\) rotated through 90° eg3 Rotation through 180° centre the origin so the shape turns
11 \(\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} a & b \\ 0 & 0.4 \end{pmatrix} = k\,\mathbf{I} \qquad\) where \(k\) is a constant and \(\mathbf{I}\) is the identity matrix.
Work out the values of \(a\) and \(b\). [4 marks]
Mark scheme
Answer
Mark
Comments
\(\begin{bmatrix} 2a & 2b + 0.4 \\ 0 & 1.2 \end{bmatrix}\) or \(2a = k\) or \(k = 1.2\) or \(2b + 0.4 = 0\)
M1
oe any 3 terms correct in correct position could be implied from second M mark
for B5, products must be seen in correct order and results of products must be correct B4 a B5 response with no indication that \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) is the identity matrix B3 (reflection in \(y = -x\)) \(\ \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) and (rotation) \(\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) and (reflection in \(x\)-axis) \(\ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) B2 Any two of the above B1 Any one of the above
for B5, products must be seen in correct order and results of products must be correct B4 a B5 response with no indication that \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) is the identity matrix B3 (reflection in \(y = -x\)) \(\ \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) and (rotation) \(\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) and (reflection in \(x\)-axis) \(\ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) B2 Any two of the above B1 Any one of the above
Alternative method 3 transforms a general point
\(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y \\ -x \end{pmatrix}\) and \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} -y \\ -x \end{pmatrix} = \begin{pmatrix} x \\ -y \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} x \\ -y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) and indication that \(\begin{pmatrix} x \\ y \end{pmatrix}\) has mapped to itself
B5
for B5, products must be seen in correct order and results of products must be correct B4 a B5 response with no indication that \(\begin{pmatrix} x \\ y \end{pmatrix}\) has mapped to itself B3 (reflection in \(y = -x\)) \(\ \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) and (rotation) \(\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) and (reflection in \(x\)-axis) \(\ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) B2 Any two of the above B1 Any one of the above
columns in 2 by 4 matrices can be in any order for B5, products must be seen in correct order and results of products must be correct B4 a B5 response with no indication that unit square has mapped to itself B3 (reflection in \(y = -x\)) \(\ \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) and (rotation) \(\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) and (reflection in \(x\)-axis) \(\ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) B2 Any two of the above B1 Any one of the above
Additional guidance
For B3, B2 and B1 the matrices must not be the answers to a product
Must use matrix multiplication
Ignore commas and lines within matrices
Allow missing brackets if arrays are correct
Examples of indication Alt 1 or 2 \(\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathrm{I}\ \) or \(\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\ \) which is the identity matrix Alt 3 \(\ \begin{pmatrix} x \\ y \end{pmatrix}\ \) which is the same as the original Alt 4 \(\ \begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}\ \) which is the same as the original
Alts 1 and 2 Indications may be seen at the start of the solution
Alt 3 \(\ \begin{pmatrix} x \\ y \end{pmatrix}\ \) must be algebraic
10 The transformation matrix \(\ \begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix}\ \) maps the point \((1,\ -3)\) onto the point \((1,\ 4)\)
Work out the values of \(a\) and \(b\). You must show your working. [5 marks]
Mark scheme
Answer
Mark
Comments
\(\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix}\begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\)
M1
oe implied by a correct equation
\(a - 3b = 1\)
A1
oe may be implied by correct answers
\(2a - 9b = 4\)
A1
Correct elimination of a variable from their 2 linear equations with both equations having the same two variables
M1
eg \(3a - 2a = 3 - 4\) or \(-6b - -9b = 2 - 4\)
\(a = -1 \qquad b = -\dfrac{2}{3}\)
A1
must be exact values
Additional guidance
\(\begin{pmatrix} a - 3b \\ 2a - 9b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) implies M1 but does not imply any correct equations
If the same method is used for both \(a\) and \(b\) (eg equates coefficients and eliminates a variable), mark the attempt that favours the student
Ignore commas and lines within matrices
Allow missing brackets if arrays are correct dimensions
Answers only \(\ a = -1\) and \(b = -\dfrac{2}{3}\)
5 marks
Allow use of \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) as a misread If solution \(a = -1\) and \(b = \dfrac{2}{3}\)
M1A0A0 M1A1ft (A1ft after misread)
Transposing \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\) is not a misread (could still score 2nd M1)
\(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) followed by \(a - 3b = 1\) and \(2a - 9b = 4\) (could still score 2nd M1)
M1A1A1
\(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) with only one of \(a - 3b = 1\) and \(2a - 9b = 4\) (could still score 2nd M1)
M1A1A0
\(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) with neither equation correct # (could still score 2nd M1)
B1 for any two or three correct elements in the correct position in a \(2 \times 2\) matrix
Additional guidance
Correct answer followed by further work, eg \(\begin{pmatrix} 20 \\ 9 \end{pmatrix}\) scores B1 only Matrices multiplied the wrong way round can score SC1 if correct \(\mathrm{BA} = \begin{pmatrix} 14 & -14 \\ 7 & 7 \end{pmatrix}\) Condone no brackets around the numbers in their \(2 \times 2\) matrix Ignore any commas that appear in their \(2 \times 2\) matrix Do not follow through on any misreads of the numbers in the given matrices
20 Under the transformation represented by \(\ \begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\ \) the image of point \(P(a, 2)\) is point \(Q\).
Can point \(Q\) be the same as point \(P\)?
You must show your working. [4 marks]
Mark scheme
Answer
Mark
Comments
\(\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} a \\ 2 \end{pmatrix}\) or \(\begin{pmatrix} -a - 6 \\ 2a + 8 \end{pmatrix}\)
M1
Allow \((-a - 6 \quad 2a + 8)\)
\(-a - 6 = a\) or \(2a + 8 = 2\)
M1
oe linear equation(s) (not \(a = -3\)) Implies M1 M1
\(-a - 6 = a\) and \(2a + 8 = 2\)
A1
oe equations (not \(a = -3\))
Shows both equations have a common solution (\(a = -3\)) and Yes
A1ft
ft M1 M1 A0 Must show that their two linear equations do not have a common solution and No SC4 \(\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}\) and Yes SC3 \(\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}\)
Additional guidance
\(\begin{pmatrix} a \\ 2 \end{pmatrix}\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\) is first M0 unless recovered
In matrices, allow missing brackets or inclusion of ‘fraction’ lines
Only one equation can score a maximum of M1 M1 A0 A0
\(a = -3\) with no correct working
Zero
\(\begin{pmatrix} -a - 6 \\ 2a + 8 \end{pmatrix} = \begin{pmatrix} a \\ 2 \end{pmatrix}\) with no further valid work
M1 M0 A0 A0
The final A mark may be seen in various ways eg1 Solves both equations obtaining \(a = -3\) each time and Yes (or shows that both equations simplify to \(2a = -6\) and Yes) eg2 Solves one equation obtaining \(a = -3\) and shows by substitution that \(a = -3\) satisfies the other equation and Yes eg3 Adds the two equations to obtain a correct statement and Yes \(\quad \begin{aligned} -2a - 6 &= 0 \\ \underline{2a + 8} &\underline{{}= 2} \\ 2 &= 2 \end{aligned}\)
