| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Transformation mapping problems |
| Difficulty | Standard +0.3 Part (i) requires computing a 3×3 determinant and showing it's never zero—straightforward algebra with no conceptual difficulty. Part (ii) involves applying a matrix transformation to two points and using the distance formula, requiring multiple steps but all standard techniques. This is slightly easier than average as it's methodical computation without requiring geometric insight or novel problem-solving. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03h Determinant 2x2: calculation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k(-20-2k) + 2(-12-2k) + 7(-3k+5k)\) or equivalent cofactor expansion | M1 | Correct method to find determinant, condone a single sign slip but second term must be \(+2\) |
| \(-2k^2 - 10k - 24 (= 0)\) isw | A1 | Correct simplified determinant |
| \(b^2 - 4ac = (10)^2 - 4(-2)(-24) = \ldots\) or completing the square: \(k^2+5k+12 \Rightarrow (k+2.5)^2 + 5.75 \Rightarrow (k+2.5)^2 = -5.75\) or differentiating: \(\frac{d(-2k^2-10k-24)}{dk} = -4k-10 = 0 \Rightarrow k = -2.5\) or quadratic formula: \(k = \frac{10 \pm \sqrt{(-10)^2 - 4(-2)(-25)}}{2(-2)} = \frac{-5 \pm \sqrt{23}\,i}{2}\) | M1 | Either: finds discriminant or sufficient working to identify sign; completes square so \((k\pm a)^2 = -b\); completes square and states \((k\pm a)^2 \geq 0\); states \(-\alpha(k\pm a)^2 \leq 0\); differentiates to find vertex; uses quadratic formula to find imaginary roots |
| \(b^2 - 4ac = -92 < 0\) therefore no real roots so non-singular; or \((k+2.5)^2 + 5.75 > 0\) therefore no real roots so non-singular; or imaginary roots therefore no real roots so non-singular | A1 | Correct solution only. Must conclude no real roots and non-singular |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}2 & -1\\-3 & 0\end{pmatrix}\begin{pmatrix}a & 4\\2 & -a\end{pmatrix} = \begin{pmatrix}\cdots & \cdots\\\cdots & \cdots\end{pmatrix}\) | M1 | Uses matrix \(\mathbf{Q}\) to find coordinates of points \(A'\) and \(B'\). Condone a sign slip |
| \(\begin{pmatrix}2a-2 & 8+a\\-3a & -12\end{pmatrix}\) or \((2a-2,\,-3a)\) and \((8+a,\,-12)\) | A1 | Correct coordinates for points \(A'\) and \(B'\), do not need to be labelled |
| \(\sqrt{\left[(2a-2)-(8+a)\right]^2 + \left[-3a-(-12)\right]^2} = \sqrt{58}\) or \(\overrightarrow{A'B'} = \binom{10-a}{-12+3a}\) | M1 | Finds distance between \(A'\) and \(B'\), must not be equal to \(A\) and \(B\), sets equal to \(\sqrt{58}\), forms a 3TQ |
| \((a-10)^2 + (12-3a)^2 = 58\) leading to \(10a^2 - 92a + 186 = 0\) | A1 | Correct 3TQ from correct coordinates |
| \(a = 3,\, \dfrac{31}{5}\) o.e. cso | A1 | Correct values |
# Question 9:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k(-20-2k) + 2(-12-2k) + 7(-3k+5k)$ or equivalent cofactor expansion | M1 | Correct method to find determinant, condone a single sign slip but second term must be $+2$ |
| $-2k^2 - 10k - 24 (= 0)$ isw | A1 | Correct simplified determinant |
| $b^2 - 4ac = (10)^2 - 4(-2)(-24) = \ldots$ or completing the square: $k^2+5k+12 \Rightarrow (k+2.5)^2 + 5.75 \Rightarrow (k+2.5)^2 = -5.75$ or differentiating: $\frac{d(-2k^2-10k-24)}{dk} = -4k-10 = 0 \Rightarrow k = -2.5$ or quadratic formula: $k = \frac{10 \pm \sqrt{(-10)^2 - 4(-2)(-25)}}{2(-2)} = \frac{-5 \pm \sqrt{23}\,i}{2}$ | M1 | Either: finds discriminant or sufficient working to identify sign; completes square so $(k\pm a)^2 = -b$; completes square and states $(k\pm a)^2 \geq 0$; states $-\alpha(k\pm a)^2 \leq 0$; differentiates to find vertex; uses quadratic formula to find imaginary roots |
| $b^2 - 4ac = -92 < 0$ therefore **no real roots** so **non-singular**; or $(k+2.5)^2 + 5.75 > 0$ therefore **no real roots** so **non-singular**; or imaginary roots therefore **no real roots** so **non-singular** | A1 | Correct solution only. Must conclude no real roots and non-singular |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}2 & -1\\-3 & 0\end{pmatrix}\begin{pmatrix}a & 4\\2 & -a\end{pmatrix} = \begin{pmatrix}\cdots & \cdots\\\cdots & \cdots\end{pmatrix}$ | M1 | Uses matrix $\mathbf{Q}$ to find coordinates of points $A'$ and $B'$. Condone a sign slip |
| $\begin{pmatrix}2a-2 & 8+a\\-3a & -12\end{pmatrix}$ or $(2a-2,\,-3a)$ and $(8+a,\,-12)$ | A1 | Correct coordinates for points $A'$ and $B'$, do not need to be labelled |
| $\sqrt{\left[(2a-2)-(8+a)\right]^2 + \left[-3a-(-12)\right]^2} = \sqrt{58}$ or $\overrightarrow{A'B'} = \binom{10-a}{-12+3a}$ | M1 | Finds distance between $A'$ and $B'$, must not be equal to $A$ and $B$, sets equal to $\sqrt{58}$, forms a 3TQ |
| $(a-10)^2 + (12-3a)^2 = 58$ leading to $10a^2 - 92a + 186 = 0$ | A1 | Correct 3TQ from correct coordinates |
| $a = 3,\, \dfrac{31}{5}$ o.e. cso | A1 | Correct values |\begin{enumerate}
\item (i)
\end{enumerate}
$$\mathbf { P } = \left( \begin{array} { r r r }
k & - 2 & 7 \\
- 3 & - 5 & 2 \\
k & k & 4
\end{array} \right)$$
where $k$ is a constant
Show that $\mathbf { P }$ is non-singular for all real values of $k$.\\
(ii)
$$\mathbf { Q } = \left( \begin{array} { r r }
2 & - 1 \\
- 3 & 0
\end{array} \right)$$
The matrix $\mathbf { Q }$ represents a linear transformation $T$\\
Under $T$, the point $A ( a , 2 )$ and the point $B ( 4 , - a )$, where $a$ is a constant, are transformed to the points $A ^ { \prime }$ and $B ^ { \prime }$ respectively.
Given that the distance $A ^ { \prime } B ^ { \prime }$ is $\sqrt { 58 }$, determine the possible values of $a$.
\hfill \mbox{\textit{Edexcel CP AS 2023 Q9 [9]}}