A2 October 2021 Paper 2 Q2
2.
\[\mathbf{A} = \begin{pmatrix} 4 & -2 \\ 5 & 3 \end{pmatrix}\]The matrix \(\mathbf{A}\) represents the linear transformation \(M\).
Prove that, for the linear transformation \(M\), there are no invariant lines.
(5)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 4 & -2 \\ 5 & 3 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} X \\ mX + c \end{pmatrix}\) leading to an equation in \(x\), \(m\), \(c\) and \(X\) | M1 | 3.1a |
| \(4x - 2(mx + c) = X\) and \(5x + 3(mx + c) = mX + c\) | A1 | 1.1b |
| \(5x + 3(mx + c) = m\left(4x - 2(mx + c)\right) + c\) leading to \(5 + 3m = 4m - 2m^2\) \(\quad\left(3c = -2mc + c\right)\) | M1 | 2.1 |
| \(2m^2 - m + 5 = 0 \Rightarrow b^2 - 4ac = (-1)^2 - 4(2)(5) = \ldots\) or Solves \(3c = -2mc + c \Rightarrow m = \ldots\) | dM1 | 1.1b |
| Correct expression for the discriminant \(= \{-39\} \lt 0\) therefore there are no invariant lines. or \(m = -1\) and shows a contradiction in \(5 + 3m = 4m - 2m^2\) therefore there are no invariant lines. | A1 | 2.4 |
| (5) |
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 4 & -2 \\ 5 & 3 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ mX \end{pmatrix}\) leading to an equation in \(x\), \(m\) and \(X\) | M1 | 3.1a |
| \(4x - 2(mx) = X\) and \(5x + 3(mx) = mX\) | A1 | 1.1b |
| \(5x + 3(mx) = m\left(4x - 2(mx)\right)\) leading to \(5 + 3m = 4m - 2m^2\) | M1 | 2.1 |
| \(2m^2 - m + 5 = 0 \Rightarrow b^2 - 4ac = (-1)^2 - 4(2)(5) = \ldots\) | dM1 | 1.1b |
| Correct expression for the discriminant \(= \{-39\} \lt 0\) therefore there are no invariant lines that pass through the origin / no invariant lines. | A1 | 2.4 |
| (5) | ||
| (5 marks) |
Notes
M1: Sets up a matrix equation in an attempt to find a fixed line and extract at least one equation.
A1: Correct equations.
M1: Eliminates \(X\) from the simultaneous equations and equates the coefficients of \(x\) leading to a quadratic equation in terms of \(m\).
dM1: Dependent on the previous method, finds the value of the discriminant, this can be seen in an attempt to solve the quadratic using the formula. Alternatively solves \(3c = -2mc + c\) and finds a value for \(m\).
Note: If the quadratic equation in \(m\) is solved on a calculator and complex roots given this is M0 as they are not showing why there are no real roots.
A1: Correct expression for the discriminant, states \(\lt 0\) and draws the required conclusion. Alternatively, correct value for \(m\), shows a contradiction in \(5 + 3m = 4m - 2m^2\) and draws the required conclusion.
Alternative
M1: Sets up a matrix equation in an attempt to find a fixed line and extract at least one equation.
A1: Correct equations.
M1: Eliminates \(X\) from the simultaneous equations and equates the coefficients of \(x\) leading to a quadratic equation in terms of \(m\).
dM1: Dependent on the previous method, finds the value of the discriminant.
A1: Correct expression for the discriminant, states \(\lt 0\) and draws the required conclusion.