| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Transformation mapping problems |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring matrix inversion using cofactors/adjugate method and understanding that inverse transformations map lines backwards. Part (a) is routine matrix inversion with a parameter. Part (b) requires recognizing that B^{-1} maps l_2 back to l_1, extracting a point and direction from the vector product form, then applying the inverse transformation. While multi-step, these are standard techniques for Further Maths students with no novel insight required. |
| Spec | 4.03o Inverse 3x3 matrix4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{A} = \begin{pmatrix}1 & -1 & 1\\1 & 1 & 1\\1 & 2 & a\end{pmatrix}\), \( | \mathbf{A} | = a - 2 + a - 1 + 2 - 1 = 2a - 2\) |
| Applies correct method to reach at least a matrix of cofactors, 2 correct rows or 2 correct columns needed | M1 | |
| Correct transpose of cofactors: \(\begin{pmatrix}a-2 & a+2 & -2\\1-a & a-1 & 0\\1 & -3 & 2\end{pmatrix}\) | A1 | |
| \(\mathbf{A}^{-1} = \frac{1}{2a-2}\begin{pmatrix}a-2 & a+2 & -2\\1-a & a-1 & 0\\1 & -3 & 2\end{pmatrix}\) | A1 | Correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 4 \Rightarrow \mathbf{A}^{-1} = \frac{1}{6}\begin{pmatrix}2 & 6 & -2\\-3 & 3 & 0\\1 & -3 & 2\end{pmatrix}\) | B1ft | Correct inverse (follow through from (a)) |
| \(\frac{1}{6}\begin{pmatrix}2 & 6 & -2\\-3 & 3 & 0\\1 & -3 & 2\end{pmatrix}\begin{pmatrix}12-6\lambda\\4+2\lambda\\6+3\lambda\end{pmatrix} = \ldots\) | M1 | Attempt to multiply the parametric form of \(l_2\) by their inverse |
| \(= \begin{pmatrix}6-\lambda\\-4+4\lambda\\2-\lambda\end{pmatrix}\) | A1 | Correct parametric form |
| \(\mathbf{r} = \begin{pmatrix}6\\-4\\2\end{pmatrix} + \lambda\begin{pmatrix}-1\\4\\-1\end{pmatrix}\) | A1 | Correct equation (allow equivalent forms) but if given as \(l = \ldots\) award A0 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{A} = \begin{pmatrix}1 & -1 & 1\\1 & 1 & 1\\1 & 2 & a\end{pmatrix}$, $|\mathbf{A}| = a - 2 + a - 1 + 2 - 1 = 2a - 2$ | B1 | Correct determinant in any form |
| Applies correct method to reach at least a matrix of cofactors, 2 correct rows or 2 correct columns needed | M1 | |
| Correct transpose of cofactors: $\begin{pmatrix}a-2 & a+2 & -2\\1-a & a-1 & 0\\1 & -3 & 2\end{pmatrix}$ | A1 | |
| $\mathbf{A}^{-1} = \frac{1}{2a-2}\begin{pmatrix}a-2 & a+2 & -2\\1-a & a-1 & 0\\1 & -3 & 2\end{pmatrix}$ | A1 | Correct inverse |
**Total: (4)**
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 4 \Rightarrow \mathbf{A}^{-1} = \frac{1}{6}\begin{pmatrix}2 & 6 & -2\\-3 & 3 & 0\\1 & -3 & 2\end{pmatrix}$ | B1ft | Correct inverse (follow through from (a)) |
| $\frac{1}{6}\begin{pmatrix}2 & 6 & -2\\-3 & 3 & 0\\1 & -3 & 2\end{pmatrix}\begin{pmatrix}12-6\lambda\\4+2\lambda\\6+3\lambda\end{pmatrix} = \ldots$ | M1 | Attempt to multiply the parametric form of $l_2$ by their inverse |
| $= \begin{pmatrix}6-\lambda\\-4+4\lambda\\2-\lambda\end{pmatrix}$ | A1 | Correct parametric form |
| $\mathbf{r} = \begin{pmatrix}6\\-4\\2\end{pmatrix} + \lambda\begin{pmatrix}-1\\4\\-1\end{pmatrix}$ | A1 | Correct equation (allow equivalent forms) but if given as $l = \ldots$ award A0 |
**Total: (4) — Question Total: 8**
---6.
$$\mathbf { A } = \left( \begin{array} { r r r }
1 & - 1 & 1 \\
1 & 1 & 1 \\
1 & 2 & a
\end{array} \right) \quad a \neq 1$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { A } ^ { - 1 }$ in terms of $a$.\\
.
The straight line $l _ { 1 }$ is mapped onto the straight line $l _ { 2 }$ by the transformation represented by the matrix $\mathbf { B }$.
$$\mathbf { B } = \left( \begin{array} { r r r }
1 & - 1 & 1 \\
1 & 1 & 1 \\
1 & 2 & 4
\end{array} \right)$$
The equation of $l _ { 2 }$ is
$$( \mathbf { r } - ( 12 \mathbf { i } + 4 \mathbf { j } + 6 \mathbf { k } ) ) \times ( - 6 \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } ) = \mathbf { 0 }$$
\item Find a vector equation for the line $l _ { 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2020 Q6 [8]}}