Question 2 - A-Level Maths

Edexcel CP2 2021 June — Question 2 5 marks

Exam Board	Edexcel
Module	CP2 (Core Pure 2)
Year	2021
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find invariant lines through origin
Difficulty	Standard +0.8 This question requires students to find eigenvalues by solving det(A - λI) = 0, which yields a quadratic with no real solutions (discriminant < 0), then conclude no invariant lines exist. While the calculation is straightforward, the conceptual link between complex eigenvalues and absence of invariant lines, plus the proof structure requirement, makes this moderately harder than average A-level questions.
Spec	4.03h Determinant 2x2: calculation

2. $$A = \left( \begin{array} { r r } 4 & - 2 \\ 5 & 3 \end{array} \right)$$ The matrix $\mathbf { A }$ represents the linear transformation $M$.
Prove that, for the linear transformation $M$, there are no invariant lines.
(5)

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Question 2:

Answer	Marks	Guidance
Answer	Mark	Guidance
$\begin{pmatrix}4 & -2\\5 & 3\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix} = \begin{pmatrix}X\\mX+c\end{pmatrix}$ leading to equation in $x$, $m$, $c$ and $X$	M1	Sets up matrix equation to find fixed line; extracts at least one equation
$4x - 2(mx+c) = X$ and $5x + 3(mx+c) = mX + c$	A1	Correct equations
$5x + 3(mx+c) = m(4x - 2(mx+c)) + c$ leading to $5 + 3m = 4m - 2m^2$ (and $3c = -2mc + c$)	M1	Eliminates $X$, equates coefficients of $x$, reaching quadratic in $m$
$2m^2 - m + 5 = 0 \Rightarrow b^2 - 4ac = (-1)^2 - 4(2)(5) = \ldots$	dM1	Dependent on previous M; finds discriminant in attempt to solve quadratic
Discriminant $= \{-39\} < 0$, therefore no invariant lines	A1	Correct discriminant expression, states $< 0$, draws required conclusion. Alternatively: $m = -1$ shows contradiction in $5 + 3m = 4m - 2m^2$, draws required conclusion

Note: If quadratic in $m$ solved on calculator giving complex roots, award M0 as method not shown.

# Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}4 & -2\\5 & 3\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix} = \begin{pmatrix}X\\mX+c\end{pmatrix}$ leading to equation in $x$, $m$, $c$ and $X$ | M1 | Sets up matrix equation to find fixed line; extracts at least one equation |
| $4x - 2(mx+c) = X$ and $5x + 3(mx+c) = mX + c$ | A1 | Correct equations |
| $5x + 3(mx+c) = m(4x - 2(mx+c)) + c$ leading to $5 + 3m = 4m - 2m^2$ (and $3c = -2mc + c$) | M1 | Eliminates $X$, equates coefficients of $x$, reaching quadratic in $m$ |
| $2m^2 - m + 5 = 0 \Rightarrow b^2 - 4ac = (-1)^2 - 4(2)(5) = \ldots$ | dM1 | Dependent on previous M; finds discriminant in attempt to solve quadratic |
| Discriminant $= \{-39\} < 0$, therefore no invariant lines | A1 | Correct discriminant expression, states $< 0$, draws required conclusion. Alternatively: $m = -1$ shows contradiction in $5 + 3m = 4m - 2m^2$, draws required conclusion |

**Note:** If quadratic in $m$ solved on calculator giving complex roots, award M0 as method not shown.

Show LaTeX source

2.

$$A = \left( \begin{array} { r r } 
4 & - 2 \\
5 & 3
\end{array} \right)$$

The matrix $\mathbf { A }$ represents the linear transformation $M$.\\
Prove that, for the linear transformation $M$, there are no invariant lines.\\
(5)

\hfill \mbox{\textit{Edexcel CP2 2021 Q2 [5]}}

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Q1 5 Q2 5 Q3 6 Q4 9 Q5 8 Q6 14 Q7 9 Q8 11 Q9 8

Edexcel CP2 2021 June — Question 2 5 marks

Question 2:

Note: If quadratic in \(m\) solved on calculator giving complex roots, award M0 as method not shown.

This paper (9 questions)