A2 June 2023 Paper 2 Q3
3.
\[\mathbf{M} = \begin{pmatrix}-2 & 5\\ 6 & k\end{pmatrix}\]where \(k\) is a constant.
Given that
\[\mathbf{M}^2 + 11\mathbf{M} = a\mathbf{I}\]where \(a\) is a constant and \(\mathbf{I}\) is the \(2 \times 2\) identity matrix,
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^2 + 11\mathbf{M} = \begin{pmatrix}a & 0\\ 0 & a\end{pmatrix} \Rightarrow \begin{pmatrix}34 & 5k - 10\\ 6k - 12 & k^2 + 30\end{pmatrix} + \begin{pmatrix}-22 & 55\\ 66 & 11k\end{pmatrix} = \begin{pmatrix}a & 0\\ 0 & a\end{pmatrix}\) | M1 | 1.1b |
| \(\Rightarrow a = 12\) | A1 | 2.2a |
| \(5k - 10 + 55 = 0 \Rightarrow 5k = -45 \Rightarrow k = -9\,*\) or \(6k - 12 + 66 = 0 \Rightarrow 6k = -54 \Rightarrow k = -9\,*\) or \(k^2 + 11k + 30 = 12 \Rightarrow k^2 + 11k + 18 = 0 \Rightarrow k = -2, -9\,*\), \(k \neq -2\) as \(5 \times -2 - 10 + 55 \neq 0\) or \(6 \times -2 - 12 + 66 \neq 0\) | A1* | 2.1 |
| (3) |
Notes
Alternative
| Scheme | Marks | AO |
|---|---|---|
| Using \(k = -9\) \(\mathbf{M}^2 + 11\mathbf{M} = \begin{pmatrix}a & 0\\ 0 & a\end{pmatrix} \Rightarrow \begin{pmatrix}34 & -55\\ -66 & 111\end{pmatrix} + \begin{pmatrix}-22 & 55\\ 66 & -99\end{pmatrix} = \begin{pmatrix}12 & 0\\ 0 & 12\end{pmatrix}\) | M1 | 1.1b |
| \(\Rightarrow a = 12\) | A1 | 2.2a |
| Conclusion: therefore \(k = -9\) | A1* | 2.1 |
| (3) |
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 \(k = -9\)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ mx + c\end{pmatrix} = \begin{pmatrix}X\\ mX + c\end{pmatrix} \Rightarrow \begin{cases}-2x + 5(mx + c) = X\\ 6x - 9(mx + c) = mX + c\end{cases}\) | M1 | 1.1b |
| \(\Rightarrow 6x - 9mx - 9c = -2mx + 5m^2x + 5mc + c\) \(\left\{\Rightarrow \left(5m^2 + 7m - 6\right)x + (5m + 10)c = 0\right\}\) | M1 A1 | 3.1a 1.1b |
| \(\Rightarrow 5m^2 + 7m - 6 = 0\ \{\Rightarrow (m + 2)(5m - 3)\} \Rightarrow m = -2, \dfrac{3}{5}\) | M1 | 1.1b |
| \(m = \dfrac{3}{5} \Rightarrow 5m + 10 \neq 0\) so need \(c = 0\) hence \(y = \dfrac{3}{5}x\) is a fixed line | A1 | 2.2a |
| \(m = -2 \Rightarrow 5m + 10 = 0\) so \(c\) can be anything, so \(y = -2x + c\) for any \(c\) is fixed. | A1 | 2.2a |
| (6) |
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
\[\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ mx\end{pmatrix} = \begin{pmatrix}X\\ mX\end{pmatrix} \Rightarrow \begin{cases}-2x + 5(mx) = X\\ 6x - 9(mx) = mX\end{cases}\]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
\[\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow \begin{cases}-2x + 5y = x\\ 6x - 9y = y\end{cases}\]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 \(y = -2x + c\)
| Scheme | Marks | AO |
|---|---|---|
| (\((0, c) \rightarrow (5c, -9c)\) so need \(c = 0\),) \((1, m) \rightarrow (-2 + 5m, 6 - 9m)\) so need or \(5m = 3\) hence \(y = \dfrac{3}{5}x\) contains fixed points. or \(\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ \frac{3}{5}x\end{pmatrix} = \begin{pmatrix}-2x + 3x\\ 6x - \frac{27}{5}x\end{pmatrix} = \begin{pmatrix}x\\ \frac{3}{5}x\end{pmatrix} \Rightarrow y = \dfrac{3}{5}x\) or \(\begin{pmatrix}-2 & 5\\ 6 & -9\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} \Rightarrow \begin{cases}-2x + 5y = x\\ 6x - 9y = y\end{cases} \Rightarrow y = \dfrac{3}{5}x\) | B1 | 3.2a |
| (1) | ||
| (10 marks) |
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.