A2 June 2022 Paper 2 Q7
7 You are given that \(a\) is a parameter which can take only real values.
The matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \begin{pmatrix} 2 & 4 & -6 \\ -3 & 10 - 4a & 9 \\ 7 & 4 & 4 \end{pmatrix}\).
You are given the following system of equations in \(x\), \(y\) and \(z\).
\[\begin{aligned} 2x + 4y - 6z &= 6 \\ -3x + (10 - 4a)y + 9z &= -9 \\ 7x + 4y + 4z &= 11 \end{aligned}\]
The system can be written in the form \(\mathbf{A}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ -9 \\ 11 \end{pmatrix}\).
The transformation represented by \(\mathbf{A}\) is denoted by T.
A 3-D object of volume \(|5a - 20|\) is transformed by T to a 3-D image.
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{A}\) \(= 2((10 - 4a) \times 4 - 9 \times 4) - 4(-3 \times 4 - 9 \times 7)\) \(\quad + (-6)(-3 \times 4 - (10 - 4a) \times 7)\) | M1 | 1.1 |
| \(= 2(40 - 16a - 36) - 4(-12 - 63)\) \(\quad - 6(-12 - 70 + 28a)\) \(= 8 - 32a + 300 + 492 - 168a\) \(= 800 - 200a\) or \(200(4 - a)\) | A1 | 1.1 |
| [2] |
Notes
M1: Expanding the determinant
Condone one calculation error
A1: ISW once all \(a\) terms and all number terms collected
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\begin{pmatrix} 4 - 16a & 75 & 28a - 82 \\ -40 & 50 & 20 \\ 96 - 24a & 0 & 32 - 8a \end{pmatrix}\) | *M1 | 1.1 |
| \(\begin{pmatrix} 4 - 16a & -40 & 96 - 24a \\ 75 & 50 & 0 \\ 28a - 82 & 20 & 32 - 8a \end{pmatrix}\) | dep*M1 | 1.1 |
| \(\mathbf{A}^{-1} = \dfrac{1}{800 - 200a}\begin{pmatrix} 4 - 16a & -40 & 96 - 24a \\ 75 & 50 & 0 \\ 28a - 82 & 20 & 32 - 8a \end{pmatrix}\) | A1FT | 1.1 |
| \(\mathbf{A}^{-1}\begin{pmatrix} 6 \\ -9 \\ 11 \end{pmatrix}\) \(= \dfrac{1}{\Delta}\begin{pmatrix} 4 - 16a & -40 & 96 - 24a \\ 75 & 50 & 0 \\ 28a - 82 & 20 & 32 - 8a \end{pmatrix}\begin{pmatrix} 6 \\ -9 \\ 11 \end{pmatrix}\) \(= \dfrac{1}{800 - 200a}\begin{pmatrix} 1440 - 360a \\ 0 \\ -320 + 80a \end{pmatrix}\) | M1 | 1.1 |
| \(= \dfrac{1}{200(4 - a)}\begin{pmatrix} 360(4 - a) \\ 0 \\ -80(4 - a) \end{pmatrix}\) \(= \dfrac{40(4 - a)}{200(4 - a)}\begin{pmatrix} 9 \\ 0 \\ -2 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 9 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} \frac{9}{5} \\ 0 \\ -\frac{2}{5} \end{pmatrix}\) So \(x = \dfrac{9}{5},\ y = 0,\ z = -\dfrac{2}{5}\) | A1 | 1.1 |
| [5] |
Notes
*M1: Correctly finding at least 5 cofactors or minors (need not be in matrix)
dep*M1: Transposing and changing signs in correct way
A1FT: FT their determinant
M1: Forming correct product and attempting multiplication, resulting in a vector
A1: Simplification to correct numerical solution (can be left as an unmultiplied vector)
i.e. A1 can be awarded for \(\begin{pmatrix} \frac{9}{5} \\ 0 \\ -\frac{2}{5} \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| (ii) Singular when \(\det\mathbf{A} = 0 \Rightarrow a = 4\) so the second equation is \(-3x - 6y + 9z = -9\)… | M1 | 2.1 |
| …which has the same normal (direction) as the first equation and is consistent with it… oe | M1 | 1.1 |
| …so the first two planes are identical and the third intersects them in a line. | A1 | 2.4 |
| [3] |
Notes
M1: Finding \(a\) from \(\det\mathbf{A} = 0\) and subbing in to 2nd equation.
Could just find the appropriate normal
M1: eg The second equation is a multiple of the first.
| Scheme | Marks | AO |
|---|---|---|
| (i) Orientation reversed if \(\det\mathbf{A} \lt 0\) so \(200(4 - a) \lt 0\) so \(a \gt 4\) | B1FT | 3.1a |
| [1] |
Notes
B1FT: Must be strict inequality
FT their expression for determinant if linear function of \(a\).
| Scheme | Marks | AO |
|---|---|---|
| (ii) Image volume smaller than object volume \(\Rightarrow -1 \lt \det\mathbf{A} \lt 1\)… | M1 | 3.1a |
| … but \(a \neq 4\) so \(\dfrac{799}{200} \lt a \lt 4\) or \(4 \lt a \lt \dfrac{801}{200}\) oe | A1 | 3.2a |
| [2] |
Notes
M1: Understanding that \(|\det\mathbf{A}|\) represents the volume scale factor and so \(|\det\mathbf{A}| \lt 1\)…
A1: …and that \(a \neq 4\).
Must be strict inequalities
\(\dfrac{799}{200} \lt a \lt \dfrac{801}{200}\) and \(a \neq 4\)