| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix inverse calculation |
| Difficulty | Standard +0.3 This is a standard Further Maths matrix question testing routine techniques: calculating a 3×3 determinant (straightforward application of formula), identifying when det=0 for singularity, and finding the inverse using the adjugate method. While it involves more computation than Core maths, these are well-practiced procedures for Further Maths students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. \(2(-a-2)-1(a-2)+2(2+2)\) (1st row) or \(2(-a-2)-1(a-4)+2(1+2)\) (1st col) | M1 | Attempt to expand determinant. Could use any row or column or other method. |
| \(=-2a-4-a+2+8=6-3a\) | A1 (AG) | Must be convincing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Matrix of cofactors: \(\begin{pmatrix}-a-2 & 2-a & 4\\ 4-a & 2a-4 & -2\\ 3 & 0 & -3\end{pmatrix}\) | M1* | At least 4 cofactors correct, or correct apart from sign. If not in matrix form only award M1 if clear where cofactors come from. Cofactor must not be multiplied by anything. Alternative method using cross product also ok. Matrix of cofactors given by \((\mathbf{C}_2\times\mathbf{C}_3, \mathbf{C}_3\times\mathbf{C}_1, \mathbf{C}_1\times\mathbf{C}_2)\) |
| A1 | 6 cofactors correct. Must include correct sign. | |
| \(\mathbf{A}^{-1}=\frac{1}{6-3a}\begin{pmatrix}-a-2 & 4-a & 3\\ 2-a & 2a-4 & 0\\ 4 & -2 & -3\end{pmatrix}\) | M1dep* | Transposing matrix of cofactors and dividing by determinant |
| A1 | cao |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $2(-a-2)-1(a-2)+2(2+2)$ (1st row) or $2(-a-2)-1(a-4)+2(1+2)$ (1st col) | M1 | Attempt to expand determinant. Could use any row or column or other method. |
| $=-2a-4-a+2+8=6-3a$ | A1 (AG) | Must be convincing |
**[2 marks total]**
---
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2$ | B1 | |
**[1 mark total]**
---
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Matrix of cofactors: $\begin{pmatrix}-a-2 & 2-a & 4\\ 4-a & 2a-4 & -2\\ 3 & 0 & -3\end{pmatrix}$ | M1* | At least 4 cofactors correct, or correct apart from sign. If not in matrix form only award M1 if clear where cofactors come from. Cofactor must not be multiplied by anything. Alternative method using cross product also ok. Matrix of cofactors given by $(\mathbf{C}_2\times\mathbf{C}_3, \mathbf{C}_3\times\mathbf{C}_1, \mathbf{C}_1\times\mathbf{C}_2)$ |
| | A1 | 6 cofactors correct. Must include correct sign. |
| $\mathbf{A}^{-1}=\frac{1}{6-3a}\begin{pmatrix}-a-2 & 4-a & 3\\ 2-a & 2a-4 & 0\\ 4 & -2 & -3\end{pmatrix}$ | M1dep* | Transposing matrix of cofactors and dividing by determinant |
| | A1 | cao |
**[4 marks total]**
---4 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { r r r } 2 & 1 & 2 \\ 1 & - 1 & 1 \\ 2 & 2 & a \end{array} \right)$.\\
(i) Show that $\operatorname { det } \mathbf { A } = 6 - 3 a$.\\
(ii) State the value of $a$ for which $\mathbf { A }$ is singular.\\
(iii) Given that $\mathbf { A }$ is non-singular find $\mathbf { A } ^ { - 1 }$ in terms of $a$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2018 Q4 [7]}}