AS June 2024 Paper 1 Q14
14 The matrix \(\mathbf{M}\) represents the transformation T, and is given by
\[\mathbf{M} = \begin{bmatrix} 3 & -1 \\ -2 & 6 \end{bmatrix}\](a) The point \(A\) has coordinates \((4, -5)\)
Find the coordinates of the image of \(A\) under T [2 marks]
(b) Show that the only invariant point under T is the origin. [3 marks]
(c) The line \(L_1\) has equation \(y = x + 1\)
The transformation T maps the line \(L_1\) onto the line \(L_2\)
Find the equation of \(L_2\) in the form \(y = mx + c\) [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Multiplies the position vector of \(A\) by \(\mathbf{M}\) to achieve at least one correct calculation or coordinate. | M1 | 1.1a |
| Obtains \((17, -38)\) Do not accept a position vector. | A1 | 1.1b |
| (2) |
Typical solution
\[\begin{aligned}\begin{bmatrix} 3 & -1 \\ -2 & 6 \end{bmatrix}\begin{bmatrix} 4 \\ -5 \end{bmatrix} &= \begin{bmatrix} 3 \times 4 + -1 \times -5 \\ -2 \times 4 + 6 \times -5 \end{bmatrix} \\ &= \begin{bmatrix} 17 \\ -38 \end{bmatrix}\end{aligned}\]Image of \(A\) is \((17, -38)\)
| Scheme | Marks | AO |
|---|---|---|
| Considers a general point \((x, y)\) Multiplies \(\mathbf{M}\begin{bmatrix} x \\ y \end{bmatrix}\) to form at least one correct equation. May be unsimplified. Or Shows that the origin is invariant, eg writes \(\mathbf{M}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) eg states that the origin is always invariant under a linear transformation. | M1 | 3.1a |
| Forms two different correct equations in \(x\) and \(y\) | M1 | 1.1a |
| Completes a reasoned argument and concludes that the origin is the only invariant point. | R1 | 2.1 |
| (3) |
Typical solution
\[\begin{bmatrix} 3 & -1 \\ -2 & 6 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}\]\[3x - y = x \quad \text{and} \quad -2x + 6y = y\]\[y = 2x \quad \text{and} \quad 2x = 5y\]\[x = 0 \quad \text{and} \quad y = 0\]So \((0, 0)\) is the only invariant point
| Scheme | Marks | AO |
|---|---|---|
| Writes the product \(\mathbf{M}\begin{bmatrix} x \\ x + 1 \end{bmatrix}\) and equates to \(\begin{bmatrix} X \\ mX + c \end{bmatrix}\) or equivalent. PI by one correct equation. Or Multiplies \(\mathbf{M}\) by a specific point on the line \(y = x + 1\) | M1 | 3.1a |
| Obtains two correct equations from \(\mathbf{M}\begin{bmatrix} x \\ x + 1 \end{bmatrix} = \begin{bmatrix} X \\ mX + c \end{bmatrix}\) or equivalent. PI Or Obtains a correct specific point on the line \(y = 2x + 8\) | A1 | 1.1b |
| Forms a correct equation in either \(m\) or \(c\) FT their two equations Or Multiplies \(\mathbf{M}\) by a second specific point on the line \(y = x + 1\) | M1 | 1.1a |
| Solves two equations in \(m\) and \(c\) Accept one incorrect equation if it has clearly come from \(\mathbf{M}\begin{bmatrix} x \\ x + 1 \end{bmatrix} = \begin{bmatrix} X \\ mX + c \end{bmatrix}\) or equivalent. Or Forms an unsimplified Cartesian equation for a straight line connecting their two points. Accept one incorrect point if it has clearly come from an attempt to find a point on the new line. | M1 | 1.1a |
| Obtains \(y = 2x + 8\) | A1 | 2.2a |
| (5) | ||
| (10 marks) |
Typical solution
\[\begin{bmatrix} 3 & -1 \\ -2 & 6 \end{bmatrix}\begin{bmatrix} x \\ x + 1 \end{bmatrix} = \begin{bmatrix} X \\ mX + c \end{bmatrix}\]\[3x - 1(x + 1) = X\]\[\text{and} \quad -2x + 6(x + 1) = mX + c\]\[2x - 1 = X \quad \text{and} \quad 4x + 6 = mX + c\]\[\text{So} \quad 4x + 6 \equiv m(2x - 1) + c\]\[4 = 2m \quad \text{and} \quad 6 = -m + c\]\[m = 2\]\[6 = -2 + c\]\[c = 8\]The new line is \(y = 2x + 8\)