| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix equation solving (AB = C) |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring multiple matrix operations: proving non-singularity via determinant, finding a 3×3 matrix inverse, and using the property (AB)^(-1) = B^(-1)A^(-1) to find B^(-1). While systematic, it requires careful algebraic manipulation across three connected parts and understanding of matrix inverse properties, placing it moderately above average difficulty. |
| Spec | 4.03j Determinant 3x3: calculation4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | Find value of det AB |
| A1 | Correct value 2 seen | |
| 2 marks | ||
| (ii) | M1 | Show correct process for adjoint entries |
| A1 | Obtain at least 4 correct entries in adjoint | |
| B1 | Divide by their determinant | |
| \((AB)^{-1} = \frac{1}{2}\begin{pmatrix} 0 & 3 & -1 \\ 0 & -1 & 1 \\ 2 & 6-3a & a-6 \end{pmatrix}\) | A1 | Obtain completely correct answer |
| 4 marks | ||
| (iii) EITHER | M1 | State or imply \((AB)^{-1} = B^{-1}A^{-1}\) |
| A1 | Obtain \(B^{-1} = (AB)^{-1} \times A\) | |
| M1 | Correct multiplication process seen | |
| A1 | Obtain three correct elements | |
| \(B^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ -6 & 2 & -2 \end{pmatrix}\) | A1 | All elements correct |
| 5 marks | ||
| OR | M1 | Attempt to find elements of B |
| A1 | All correct | |
| M1 | Correct process for \(B^{-1}\) | |
| A1 | 3 elements correct | |
| A1 | All elements correct |
**(i)** | M1 | Find value of det AB
| A1 | Correct value 2 seen
| **2 marks**
**(ii)** | M1 | Show correct process for adjoint entries
| A1 | Obtain at least 4 correct entries in adjoint
| B1 | Divide by their determinant
$(AB)^{-1} = \frac{1}{2}\begin{pmatrix} 0 & 3 & -1 \\ 0 & -1 & 1 \\ 2 & 6-3a & a-6 \end{pmatrix}$ | A1 | Obtain completely correct answer
| **4 marks**
**(iii)** EITHER | M1 | State or imply $(AB)^{-1} = B^{-1}A^{-1}$
| A1 | Obtain $B^{-1} = (AB)^{-1} \times A$
| M1 | Correct multiplication process seen
| A1 | Obtain three correct elements
$B^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ -6 & 2 & -2 \end{pmatrix}$ | A1 | All elements correct
| **5 marks**
OR | M1 | Attempt to find elements of B
| A1 | All correct
| M1 | Correct process for $B^{-1}$
| A1 | 3 elements correct
| A1 | All elements correct10 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { r r r } a & 8 & 10 \\ 2 & 1 & 2 \\ 4 & 3 & 6 \end{array} \right)$. The matrix $\mathbf { B }$ is such that $\mathbf { A B } = \left( \begin{array} { l l l } a & 6 & 1 \\ 1 & 1 & 0 \\ 1 & 3 & 0 \end{array} \right)$.\\
(i) Show that $\mathbf { A B }$ is non-singular.\\
(ii) Find $( \mathbf { A B } ) ^ { - 1 }$.\\
(iii) Find $\mathbf { B } ^ { - 1 }$.
\hfill \mbox{\textit{OCR FP1 2008 Q10 [11]}}