| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Matrix powers by induction |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard induction technique for matrix powers (base case, inductive step with matrix multiplication) followed by routine application of determinant for area scaling and substitution into the given formula. The induction follows a predictable pattern for upper triangular matrices, and part (b) involves simple algebraic manipulation. Slightly easier than average due to the mechanical nature of the tasks. |
| Spec | 4.01a Mathematical induction: construct proofs4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n=1 \Rightarrow \mathbf{M}^1 = \begin{pmatrix} 3^1 & \frac{a}{2}(3^1-1) \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}\) | B1 | Must see substitution in RHS minimum; reaches \(\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}\) |
| Assume true for \(n = k\): \(\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}^k = \begin{pmatrix} 3^k & \frac{a}{2}(3^k-1) \\ 0 & 1 \end{pmatrix}\) | M1 | Assumes result true for \(n = k\) |
| \(\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 3^k & \frac{a}{2}(3^k-1) \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}\) | M1 | Sets up matrix multiplication of assumed result by original matrix, either way round |
| \(= \begin{pmatrix} 3(3^k) & a(3^k)+\frac{a}{2}(3^k-1) \\ 0 & 1 \end{pmatrix}\) | A1 | Achieves a correct un-simplified matrix |
| \(= \begin{pmatrix} 3^{k+1} & \frac{a}{2}[2(3^k)+(3^k-1)] \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3^{k+1} & \frac{a}{2}[3(3^k)-1] \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3^{k+1} & \frac{a}{2}(3^{k+1}-1) \\ 0 & 1 \end{pmatrix}\) | A1 | Correct simplified matrix with no errors; correct un-simplified matrix seen previously and at least one correct intermediate line |
| If true for \(n=k\) then true for \(n=k+1\), and as true for \(n=1\), true for all positive integers \(n\) | A1 | Correct conclusion conveying all four key ideas; dependent on all previous marks except B1 (but \(n=1\) must have been attempted) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\det(\mathbf{M}^n) = 3^n\) or \(\det(\mathbf{M}) = 3\) | B1 | States correct determinant; may be implied by correct equation |
| \(5 \times \det(\mathbf{M}^n) = 1215 \Rightarrow p^n = q \Rightarrow n = \ldots\); \(5 \times 3^n = 1215 \Rightarrow 3^n = 243 \Rightarrow n = \ldots\) | M1 | Correct method using \(5 \times \det(\mathbf{M}^n) = 1215\), solving index equation \(p^n = q\) where \(n > 1\) |
| \(n = 5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} 3^n & \frac{a}{2}(3^n-1) \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 123 \\ -2 \end{pmatrix} \Rightarrow 2(3^n) - 2\cdot\frac{a}{2}(3^n-1) = 123 \Rightarrow a = \ldots\) | M1 | Sets up equation multiplying \(\mathbf{M}^n\) by \(\begin{pmatrix} 2 \\ -2 \end{pmatrix}\) setting equal to \(\begin{pmatrix} 123 \\ -2 \end{pmatrix}\); follow through on their \(n\) |
| \(a = 1.5\) | A1 |
## Question 3:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n=1 \Rightarrow \mathbf{M}^1 = \begin{pmatrix} 3^1 & \frac{a}{2}(3^1-1) \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}$ | B1 | Must see substitution in RHS minimum; reaches $\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}$ |
| Assume true for $n = k$: $\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}^k = \begin{pmatrix} 3^k & \frac{a}{2}(3^k-1) \\ 0 & 1 \end{pmatrix}$ | M1 | Assumes result true for $n = k$ |
| $\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 3^k & \frac{a}{2}(3^k-1) \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & a \\ 0 & 1 \end{pmatrix}$ | M1 | Sets up matrix multiplication of assumed result by original matrix, either way round |
| $= \begin{pmatrix} 3(3^k) & a(3^k)+\frac{a}{2}(3^k-1) \\ 0 & 1 \end{pmatrix}$ | A1 | Achieves a correct un-simplified matrix |
| $= \begin{pmatrix} 3^{k+1} & \frac{a}{2}[2(3^k)+(3^k-1)] \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3^{k+1} & \frac{a}{2}[3(3^k)-1] \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3^{k+1} & \frac{a}{2}(3^{k+1}-1) \\ 0 & 1 \end{pmatrix}$ | A1 | Correct simplified matrix with no errors; correct un-simplified matrix seen previously and at least one correct intermediate line |
| If true for $n=k$ then true for $n=k+1$, and as true for $n=1$, true for all positive integers $n$ | A1 | Correct conclusion conveying all four key ideas; dependent on all previous marks except B1 (but $n=1$ must have been attempted) |
### Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\det(\mathbf{M}^n) = 3^n$ or $\det(\mathbf{M}) = 3$ | B1 | States correct determinant; may be implied by correct equation |
| $5 \times \det(\mathbf{M}^n) = 1215 \Rightarrow p^n = q \Rightarrow n = \ldots$; $5 \times 3^n = 1215 \Rightarrow 3^n = 243 \Rightarrow n = \ldots$ | M1 | Correct method using $5 \times \det(\mathbf{M}^n) = 1215$, solving index equation $p^n = q$ where $n > 1$ |
| $n = 5$ | A1 | |
### Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} 3^n & \frac{a}{2}(3^n-1) \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 123 \\ -2 \end{pmatrix} \Rightarrow 2(3^n) - 2\cdot\frac{a}{2}(3^n-1) = 123 \Rightarrow a = \ldots$ | M1 | Sets up equation multiplying $\mathbf{M}^n$ by $\begin{pmatrix} 2 \\ -2 \end{pmatrix}$ setting equal to $\begin{pmatrix} 123 \\ -2 \end{pmatrix}$; follow through on their $n$ |
| $a = 1.5$ | A1 | |\begin{enumerate}
\item $\mathbf { M } = \left( \begin{array} { l l } 3 & a \\ 0 & 1 \end{array} \right) \quad$ where $a$ is a constant\\
(a) Prove by mathematical induction that, for $n \in \mathbb { N }$
\end{enumerate}
$$\mathbf { M } ^ { n } = \left( \begin{array} { c c }
3 ^ { n } & \frac { a } { 2 } \left( 3 ^ { n } - 1 \right) \\
0 & 1
\end{array} \right)$$
Triangle $T$ has vertices $A , B$ and $C$.\\
Triangle $T$ is transformed to triangle $T ^ { \prime }$ by the transformation represented by $\mathbf { M } ^ { n }$ where $n \in \mathbb { N }$
Given that
\begin{itemize}
\item triangle $T$ has an area of $5 \mathrm {~cm} ^ { 2 }$
\item triangle $T ^ { \prime }$ has an area of $1215 \mathrm {~cm} ^ { 2 }$
\item vertex $A ( 2 , - 2 )$ is transformed to vertex $A ^ { \prime } ( 123 , - 2 )$\\
(b) determine\\
(i) the value of $n$\\
(ii) the value of $a$
\end{itemize}
\hfill \mbox{\textit{Edexcel CP2 2022 Q3 [11]}}