| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix equation solving (AB = C) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring standard matrix inversion (using cofactors/adjugate method) followed by a simple matrix multiplication N = M^(-1)C. While it involves 3×3 matrices which are more computational than 2×2, the techniques are routine and well-practiced in F3. The parameter k adds minimal complexity since part (b) sets k=0. Slightly above average difficulty due to the computational load of 3×3 matrices, but no conceptual challenges or novel insights required. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \( | \mathbf{M} | = 3 - k - k(-3-1) (= 3k+3)\) |
| \(\mathbf{M}^T = \begin{pmatrix} 1 & -1 & 1 \\ k & 1 & k \\ 0 & 1 & 3 \end{pmatrix}\) or minors/cofactors matrix stated correctly | B1 | Correct transpose or minors or cofactors matrix |
| \(\mathbf{M}^{-1} = \frac{1}{3+3k}\begin{pmatrix} 3-k & -3k & k \\ 4 & 3 & -1 \\ -k-1 & 0 & 1+k \end{pmatrix}\) | M1A1ftA1ft | M1: Full attempt at inverse including reciprocal of determinant (at least 6 correct elements). A1ft: Two rows or two columns correct. A1ft: Fully correct inverse |
| NB: If every element is the negative of the correct element, allow M1A1A0 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{MN} = \begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix} \Rightarrow \mathbf{N} = \mathbf{M}^{-1}\begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix}\) | B1 | Correct statement |
| \(\mathbf{N} = \frac{1}{3}\begin{pmatrix} 3 & 0 & 0 \\ 4 & 3 & -1 \\ -1 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 5 & 6 \\ 7 & 5 & 10 \\ 0 & -1 & -3 \end{pmatrix}\) | M1A(2,1,0) | M1: Multiplies by \(\mathbf{M}^{-1}\) in correct order, must include the \(\frac{1}{3}\). A2: Correct matrix (−1 each error). If \(\frac{1}{3}\) left outside matrix award A0 |
| (4) | ||
| Total 9 |
## Question 4:
**Part (a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $|\mathbf{M}| = 3 - k - k(-3-1) (= 3k+3)$ | B1 | Correct determinant in any form |
| $\mathbf{M}^T = \begin{pmatrix} 1 & -1 & 1 \\ k & 1 & k \\ 0 & 1 & 3 \end{pmatrix}$ or minors/cofactors matrix stated correctly | B1 | Correct transpose or minors or cofactors matrix |
| $\mathbf{M}^{-1} = \frac{1}{3+3k}\begin{pmatrix} 3-k & -3k & k \\ 4 & 3 & -1 \\ -k-1 & 0 & 1+k \end{pmatrix}$ | M1A1ftA1ft | M1: Full attempt at inverse including reciprocal of determinant (at least 6 correct elements). A1ft: Two rows or two columns correct. A1ft: Fully correct inverse |
| NB: If every element is the negative of the correct element, allow M1A1A0 | | (5) |
**Part (b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{MN} = \begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix} \Rightarrow \mathbf{N} = \mathbf{M}^{-1}\begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix}$ | B1 | Correct statement |
| $\mathbf{N} = \frac{1}{3}\begin{pmatrix} 3 & 0 & 0 \\ 4 & 3 & -1 \\ -1 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 5 & 6 \\ 7 & 5 & 10 \\ 0 & -1 & -3 \end{pmatrix}$ | M1A(2,1,0) | M1: Multiplies by $\mathbf{M}^{-1}$ in correct order, must include the $\frac{1}{3}$. A2: Correct matrix (−1 each error). If $\frac{1}{3}$ left outside matrix award A0 |
| | | (4) |
| | | **Total 9** |
---4.
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & k & 0 \\
- 1 & 1 & 1 \\
1 & k & 3
\end{array} \right) , \text { where } k \text { is a constant }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { M } ^ { - 1 }$ in terms of $k$.
Hence, given that $k = 0$
\item find the matrix $\mathbf { N }$ such that
$$\mathbf { M N } = \left( \begin{array} { r r r }
3 & 5 & 6 \\
4 & - 1 & 1 \\
3 & 2 & - 3
\end{array} \right)$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q4 [9]}}