| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find inverse then solve system |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring matrix inversion (using cofactors/adjugate method for a 3×3 matrix with parameter k) and solving a linear system via matrix multiplication. The block structure (zeros in third column) simplifies calculations significantly. While more involved than basic C1-C4 content, it's a standard textbook exercise for FP3 with no novel problem-solving required. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\det\mathbf{M} = 6-k^2\) | B1 | A correct (possibly un-simplified) determinant |
| \(\mathbf{M}^T = \begin{pmatrix}3&k&k\\k&2&0\\0&0&1\end{pmatrix}\) or minors/cofactors shown | B1 | Correct transpose or minors/cofactors matrix |
| \(\frac{1}{6-k^2}\begin{pmatrix}2&-k&0\\-k&3&0\\-2k&k^2&6-k^2\end{pmatrix}\) | M1A1A1 | M1: Identifiable full attempt at inverse including reciprocal of determinant (could be indicated by at least 6 correct elements). A1: Two rows or two columns correct (ignoring determinant) — BUT M0A1A0 or M0A1A1 not possible. A1: Fully correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}a\\b\\c\end{pmatrix} = \frac{1}{5}\begin{pmatrix}2&-1&0\\-1&3&0\\-2&1&5\end{pmatrix}\begin{pmatrix}-5\\10\\7\end{pmatrix}\) | M1 | Uses \(k=1\) in the inverse and attempts to multiply to obtain a numerical value for at least one of \(a\), \(b\) or \(c\) |
| \(x=-4,\ y=7,\ z=11\) | M1A1cao | M1: Obtains values for all three coordinates. A1: Correct coordinates |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\det\mathbf{M} = 6-k^2$ | B1 | A correct (possibly un-simplified) determinant |
| $\mathbf{M}^T = \begin{pmatrix}3&k&k\\k&2&0\\0&0&1\end{pmatrix}$ or minors/cofactors shown | B1 | Correct transpose or minors/cofactors matrix |
| $\frac{1}{6-k^2}\begin{pmatrix}2&-k&0\\-k&3&0\\-2k&k^2&6-k^2\end{pmatrix}$ | M1A1A1 | M1: Identifiable full attempt at inverse **including reciprocal of determinant** (could be indicated by at least 6 correct elements). A1: Two rows or two columns correct (ignoring determinant) — **BUT M0A1A0 or M0A1A1 not possible**. A1: Fully correct inverse |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}a\\b\\c\end{pmatrix} = \frac{1}{5}\begin{pmatrix}2&-1&0\\-1&3&0\\-2&1&5\end{pmatrix}\begin{pmatrix}-5\\10\\7\end{pmatrix}$ | M1 | Uses $k=1$ in the inverse and attempts to multiply to obtain a numerical value for at least one of $a$, $b$ or $c$ |
| $x=-4,\ y=7,\ z=11$ | M1A1cao | M1: Obtains values for all three coordinates. A1: Correct coordinates |
---4. A non-singular matrix $\mathbf { M }$ is given by
$$\mathbf { M } = \left( \begin{array} { l l l }
3 & k & 0 \\
k & 2 & 0 \\
k & 0 & 1
\end{array} \right) \text {, where } k \text { is a constant. }$$
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $k$, the inverse of the matrix $\mathbf { M }$.
The point $A$ is mapped onto the point ( $- 5,10,7$ ) by the transformation represented by the matrix
$$\left( \begin{array} { l l l }
3 & 1 & 0 \\
1 & 2 & 0 \\
1 & 0 & 1
\end{array} \right)$$
\item Find the coordinates of the point $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2014 Q4 [8]}}