A2 June 2019 Paper 2 Q7
7.
\[\mathbf{M} = \begin{pmatrix}2 & -1 & 1\\ 3 & k & 4\\ 3 & 2 & -1\end{pmatrix} \qquad \text{where } k \text{ is a constant}\]| Scheme | Marks | AO |
|---|---|---|
| \(|\mathbf{M}| = 2(-k - 8) + 1(-3 - 12) + 1(6 - 3k) = 0 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k \neq -5\) | A1 | 2.4 |
| (2) |
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. \(k \lt -5,\ k \gt -5\)
Way 1
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M} = \begin{pmatrix}2 & -1 & 1\\ 3 & -6 & 4\\ 3 & 2 & -1\end{pmatrix} \Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix}p\\1\\0\end{pmatrix}\) | M1 | 3.1a |
| \(\mathbf{M}^{-1} = \dfrac{1}{5}\begin{pmatrix}-2 & 1 & 2\\ 15 & -5 & -5\\ 24 & -7 & -9\end{pmatrix}\) | B1 | 1.1b |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \dfrac{1}{5}\begin{pmatrix}-2 & 1 & 2\\ 15 & -5 & -5\\ 24 & -7 & -9\end{pmatrix}\begin{pmatrix}p\\1\\0\end{pmatrix} \Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix} = \ldots\) | M1 | 2.1 |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \dfrac{1}{5}\begin{pmatrix}-2p + 1\\15p - 5\\24p - 7\end{pmatrix}\) | A1 | 1.1b |
| \(\left(\dfrac{-2p + 1}{5},\ 3p - 1,\ \dfrac{24p - 7}{5}\right)\) | A1ft | 2.5 |
| (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
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{aligned}2x - y + z &= p\\ 3x - 6y + 4z &= 1\\ 3x + 2y - z &= 0\end{aligned} \Rightarrow \text{e.g. } \begin{aligned}8y - 5z &= -1\\ 9y - 5z &= 3p - 2\end{aligned} \Rightarrow y = \ldots\) \(\Rightarrow x = \ldots,\ z = \ldots\) | M1 | 3.1a |
| \(y = 3p - 1\) (or \(x = \dfrac{-2p + 1}{5}\) or \(z = \dfrac{24p - 7}{5}\)) | B1 | 1.1b |
| \(8(3p - 1) - 5z = -1 \Rightarrow z = \ldots \Rightarrow x = \ldots\) | M1 | 2.1 |
| \(z = \dfrac{24p - 7}{5},\ x = \dfrac{-2p + 1}{5}\) | A1 | 1.1b |
| \(\left(\dfrac{-2p + 1}{5},\ 3p - 1,\ \dfrac{24p - 7}{5}\right)\) | A1ft | 2.5 |
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 \(z\).
| Scheme | Marks | AO |
|---|---|---|
| (i) For consistency: E.g. \(5x + y = 4 - q\) and \(15x + 3y = q\) | M1 | 3.1a |
| \(4 - q = \dfrac{q}{3} \Rightarrow q = \ldots\) | M1 | 2.1 |
| \(q = 3\) | A1 | 1.1b |
| Alternative for (c)(i): \(x = 1 \Rightarrow 2 - y + z = 1,\ 3 + 2y - z = 0 \Rightarrow y = \ldots,\ z = \ldots\) M1 for allocating a number to one variable and solves for the other 2 \(x = 1,\ y = -4,\ z = -5 \Rightarrow 3 + 20 - 20 = q\) M1 substitutes into the second equation and solves for \(q\) A1: \(q = 3\) | ||
| (ii) Three planes that intersect in a line Or Three planes that form a sheaf allow sheath! | B1 | 2.4 |
| (4) | ||
| (11 marks) |
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.