AS June 2024 Paper 1 Q8
8 Three transformations, \(\mathrm{T_A}\), \(\mathrm{T_B}\) and \(\mathrm{T_C}\), are represented by the matrices \(\mathbf{A}\), \(\mathbf{B}\) and \(\mathbf{C}\) respectively.
You are given that \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\).
Transformations R and S are each defined as being the result of successive transformations, as specified in the table.
| Transformation | First transformation | followed by |
|---|---|---|
| R | \(\mathrm{T_A}\) followed by \(\mathrm{T_B}\) | \(\mathrm{T_C}\) |
| S | \(\mathrm{T_A}\) | \(\mathrm{T_B}\) followed by \(\mathrm{T_C}\) |
A quadrilateral, \(Q\), has vertices \(D\), \(E\), \(F\) and \(G\) in anticlockwise order from \(D\). Under transformation R, \(Q\)’s image, \(Q'\), has vertices \(D'\), \(E'\), \(F'\) and \(G'\) (where \(D'\) is the image of \(D\), etc). The area of \(Q\), in suitable units, is 5.
You are given that \(\det\mathbf{C} = a^2 + 1\) where \(a\) is a real constant.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A}^{-1} = \dfrac{1}{3}\begin{pmatrix} 3 & 0 \\ -2 & 1 \end{pmatrix}\) | B1 | 1.1 |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| \([\mathbf{AB} =]\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & -3 \end{pmatrix}\) | M1 | 1.1 |
| \([\mathbf{BA} =]\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -2 & -3 \end{pmatrix}\) so they are not the same | A1 | 2.2a |
| [2] |
Notes
M1: Correctly finding either \(\mathbf{AB}\) or \(\mathbf{BA}\).
A1: Must give correct calculation and correct conclusion.
\(\mathbf{AB} \ne \mathbf{BA}\) is fine for conclusion. General statement “matrices not commutable not OK unless linked to particular case.
| Scheme | Marks | AO |
|---|---|---|
| The matrix representing R is \(\mathbf{C}(\mathbf{BA})\)... | M1 | 3.1a |
| ...and the matrix representing S is \((\mathbf{CB})\mathbf{A}\) and by associativity (of matrix multiplication), \(\mathbf{C}(\mathbf{BA}) = (\mathbf{CB})\mathbf{A}\) (so R and S are the same) | A1 | 3.2a |
| [2] |
Notes
M1: Or the matrix representing S is \((\mathbf{CB})\mathbf{A}\)
Must have \(\mathbf{R} = \mathbf{C}(\mathbf{BA})\) or \(\mathbf{S} = (\mathbf{CB})\mathbf{A}\) with brackets or equivalent correct
Need to see the brackets or equivalent
BOD sight of \(\mathrm{T_A}\)
A1: Correct form for S and either explicit equality or associativity property explicitly mentioned.
M1 A0 only if state that matrices are commutative
If M0 then SC B1 for observation that \((\mathbf{AB})\mathbf{C} = \mathbf{A}(\mathbf{BC})\) or any other correct statement of associativity of matrix multiplication.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\det\mathbf{A} = 3\) or \(\det\mathbf{B} = -1\) | M1 | 1.1 |
| \(\det(\mathbf{CBA}) = -3(a^2 + 1) \lt 0\) (since \(a\) is real) [so the orientation is reversed]. Order is \(D'\), \(G'\), \(F'\), \(E'\) | A1 | 2.2a |
| [2] | ||
| (ii) \(15(a^2 + 1)\) | B1FT | 1.1 |
| [1] | ||
| (iii) The determinant is not zero... | M1 | 1.1 |
| ...so the inverse transformation exists. | A1FT | 2.2a |
| [2] |
Notes
(d)(i)
M1: Correctly calculating the determinant of A or B
Question says “Determine” so answer only is 0
A1: Some justification must be given but condone incorrect order of matrices. Orientation reversed is fine, but must see correct det \((\mathbf{CBA})\)
Can consider \(\det(\mathbf{C})\det(\mathbf{B})\det(\mathbf{A})\) instead, i.e. the effect on orientation of each of the three transformations
Allow “the order is clockwise”.
SC1 If neither det \(\mathbf{A}\) or det \(\mathbf{B}\) explicitly calculated then allow B1 for det \(\mathbf{A}\), det \(\mathbf{C}\) positive and det \(\mathbf{B}\) negative so orientation reversed.
SC2 If only det (C) considered (i.e. candidate thinks transformation R is represented by \(\mathbf{C}\)) then allow B1 for \(a^2 + 1 \gt 0\) so orientation is the same
(d)(ii)
B1FT: FT \(5 \times |\)their determinant from (i)\(|\) (if found)
FT only if their determinant is negative
(d)(iii)
M1: Understanding that the matrix associated with the transformation is non-singular.
Allow M1 for “the determinant is negative”.
Or equating det to 0 and solving even if conclusion wrong
Can be considering det \(\mathbf{C}\) or det \(\mathbf{R}\) here
Allow M1 for “determinant is positive” ONLY if it is clear where the determinant has come from (e.g. from det \(\mathbf{C}\)).
A1FT: Condone use of “matrix” rather than “transformation” (and vice versa as applicable).
Allow follow through if they are using det C.
If have found matrix singular when \(a = \mathrm{i}\) then need to discount this as not real.
If their determinant is 0 then SC1 only can be awarded for showing understanding that the transformation associated with a singular matrix does not have an inverse.
Need the be showing non-zero determinant.