AS June 2019 Paper 1 Q12
12 The matrix \(\mathbf{A}\) is given by
\[\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\](a) Prove by induction that, for all integers \(n \geqslant 1\),\[\mathbf{A}^n = \begin{bmatrix} 1 & 3^n - 1 \\ 0 & 3^n \end{bmatrix}\]
[4 marks]
(b) Find all invariant lines under the transformation matrix \(\mathbf{A}\).
Fully justify your answer. [6 marks]
Fully justify your answer. [6 marks]
(c) Find a line of invariant points under the transformation matrix \(\mathbf{A}\). [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Demonstrates the rule is correct for \(n = 1\) and states that it is true for \(n = 1\) (may appear at any stage). | B1 | 1.1b |
| Multiplies \(\begin{bmatrix} 1 & 3^k - 1 \\ 0 & 3^k \end{bmatrix}\) and \(\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\) Accept any letter in place of \(k\) (condone \(n\)). | M1 | 2.4 |
| Obtains \(\begin{bmatrix} 1 & 3^{k+1} - 1 \\ 0 & 3^{k+1} \end{bmatrix}\) from multiplying \(\begin{bmatrix} 1 & 3^k - 1 \\ 0 & 3^k \end{bmatrix}\) and \(\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\) Must include an intermediate step for the top right element. | A1 | 2.2a |
| Completes a rigorous argument and explains how their argument proves the required result. e.g. states “assume that the rule is true for \(n = k\)” (or equivalent) and “also true for \(n = k + 1\)” (or equivalent) and “for all \(n\)” and includes the base case with a conclusion. Do not accept the use of \(n\) in place of \(k\). NMS scores 0/4 | R1 | 2.1 |
Typical solution
Try \(n = 1\):
\[\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}^1 = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 1 & 3^1 - 1 \\ 0 & 3^1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\]\(\therefore\) true for \(n = 1\)
Assume true for \(n = k\)
\[\therefore \mathbf{A}^k = \begin{bmatrix} 1 & 3^k - 1 \\ 0 & 3^k \end{bmatrix}\]\[\Rightarrow \mathbf{A}^k \times \mathbf{A} = \begin{bmatrix} 1 & 3^k - 1 \\ 0 & 3^k \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\]\[\Rightarrow \mathbf{A}^{k+1} = \begin{bmatrix} 1 & 2 + 3(3^k - 1) \\ 0 & 3^k \times 3 \end{bmatrix} = \begin{bmatrix} 1 & 3^{k+1} - 1 \\ 0 & 3^{k+1} \end{bmatrix}\]\(\therefore\) it is also true for \(n = k + 1\)
True for \(n = 1\), and true for \(n = k \Rightarrow\) true for \(n = k + 1\)
Then, by induction, it is true for all integers \(n \geqslant 1\)
| Scheme | Marks | AO |
|---|---|---|
| Sets up two equations in \((x, y)\) and its image \((x\prime, y\prime)\) | M1 | 3.1a |
| Correctly substitutes \(y = mx + c\) and \(y\prime = mx\prime + c\) | A1 | 1.1b |
| Eliminates one variable to leave an equation in \(m\), \(c\) and just one other variable. | M1 | 1.1a |
| Compares coefficients to produce two correct equations in \(m\) and \(c\) | A1 | 1.1b |
| Gives \(y = 0\) or \(y = x + c\) as invariant lines. Condone other incorrect invariant lines. | B1 | 1.1b |
| Gives \(y = 0\) and \(y = x + c\) as invariant lines, with no incorrect invariant lines. NMS can score 2/6 | B1 | 2.2a |
Typical solution
\[\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x\prime \\ y\prime \end{bmatrix}\]\[x\prime = x + 2y \quad \text{and} \quad y\prime = 3y\]\[x\prime = x + 2(mx + c) \quad \text{and} \quad mx\prime + c = 3(mx + c)\]\[m(x + 2mx + 2c) + c \equiv 3mx + 3c\]\[m + 2m^2 = 3m \quad \text{and} \quad 2mc + c = 3c\]\[m(m - 1) = 0 \quad \text{and} \quad c(m - 1) = 0\]\[m = 0 \text{ or } m = 1 \quad \text{and} \quad c = 0 \text{ or } m = 1\]\[y = 0x + 0 \quad \text{or} \quad y = 1x + c\]Invariant lines are \(y = 0\) and \(y = x + c\)
Alternative mark scheme for students who assume that all invariant lines pass through the origin – max 3 marks
| Scheme | Marks | AO |
|---|---|---|
| Sets up two equations in \((x, y)\) and its image \((x\prime, y\prime)\) | M1 | 3.1a |
| Eliminates three variables to leave an equation in \(m\) and just one other variable. | M1 | 1.1a |
| Gives \(y = 0\) and \(y = x\) as invariant lines, with no other incorrect invariant lines. NMS can score 1/3 | B1 | 2.2a |
Typical solution
\[\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x\prime \\ y\prime \end{bmatrix}\]\[x\prime = x + 2y \quad \text{and} \quad y\prime = 3y\]\[x\prime = x + 2(mx) \quad \text{and} \quad mx\prime = 3(mx)\]\[m(x + 2mx) \equiv 3mx\]\[m + 2m^2 = 3m\]\[m(m - 1) = 0\]\[m = 0 \text{ or } m = 1\]\[y = 0x \quad \text{or} \quad y = 1x\]Invariant lines are \(y = 0\) and \(y = x\)
| Scheme | Marks | AO |
|---|---|---|
| Sets up at least one correct equation in \(x\) and \(y\) Accept alternative variables for this mark. | M1 | 1.1a |
| Gives \(y = 0\) as the only line of invariant points. NMS can score 2/2 | A1 | 1.1b |
| (12 marks) |