| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving matrix equations for unknown matrix |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring routine application of the 2×2 matrix inverse formula and solving AB=C for B by multiplying both sides by A^(-1). While it involves multiple steps, each step uses standard techniques with no conceptual challenges or novel insights required. |
| Spec | 4.03n Inverse 2x2 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{1}{2}\begin{pmatrix} 8 & -2 \\ -3 & 1 \end{pmatrix}\) | B1 B1 | Transpose leading diagonal and negate other diagonal; Divide by determinant |
| (ii) Either: \(\frac{1}{2}\begin{pmatrix} 14 & 2 \\ -5 & 0 \end{pmatrix}\) | B1 M1A1 M1 A1ft A1 | State or imply \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\); Use this result and obtain \(\mathbf{B}^{-1} = \mathbf{C}^{-1}\mathbf{A}\), or equivalent matrix algebra; Matrix multn., two elements correct, for any pair; All elements correct ft their (i); Correct \(\mathbf{B}^{-1}\) |
| Or: \(\mathbf{B} = \mathbf{A}^{-1}\mathbf{C}\) then \(\mathbf{B} = \frac{1}{5}\begin{pmatrix} 0 & -2 \\ 5 & 14 \end{pmatrix}\) then \(\frac{1}{2}\begin{pmatrix} 14 & 2 \\ -5 & 0 \end{pmatrix}\) | B1 M1 M1 A1A1 A1 | Find \(\mathbf{A}^{-1}\); Premultiply by \(\mathbf{A}^{-1}\) stated or implied; Matrix multn. Two elements correct; All elements correct; Solve one pair of simultaneous equations; Each pair of answers; Correct \(\mathbf{B}^{-1}\) |
(i) $\frac{1}{2}\begin{pmatrix} 8 & -2 \\ -3 & 1 \end{pmatrix}$ | B1 B1 | Transpose leading diagonal and negate other diagonal; Divide by determinant
(ii) Either: $\frac{1}{2}\begin{pmatrix} 14 & 2 \\ -5 & 0 \end{pmatrix}$ | B1 M1A1 M1 A1ft A1 | State or imply $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$; Use this result and obtain $\mathbf{B}^{-1} = \mathbf{C}^{-1}\mathbf{A}$, or equivalent matrix algebra; Matrix multn., two elements correct, for any pair; All elements correct ft their (i); Correct $\mathbf{B}^{-1}$
Or: $\mathbf{B} = \mathbf{A}^{-1}\mathbf{C}$ then $\mathbf{B} = \frac{1}{5}\begin{pmatrix} 0 & -2 \\ 5 & 14 \end{pmatrix}$ then $\frac{1}{2}\begin{pmatrix} 14 & 2 \\ -5 & 0 \end{pmatrix}$ | B1 M1 M1 A1A1 A1 | Find $\mathbf{A}^{-1}$; Premultiply by $\mathbf{A}^{-1}$ stated or implied; Matrix multn. Two elements correct; All elements correct; Solve one pair of simultaneous equations; Each pair of answers; Correct $\mathbf{B}^{-1}$
**Total: 7 marks**6 The matrix $\mathbf { C }$ is given by $\mathbf { C } = \left( \begin{array} { l l } 1 & 2 \\ 3 & 8 \end{array} \right)$.\\
(i) Find $\mathbf { C } ^ { - 1 }$.\\
(ii) Given that $\mathbf { C } = \mathbf { A B }$, where $\mathbf { A } = \left( \begin{array} { l l } 2 & 1 \\ 1 & 3 \end{array} \right)$, find $\mathbf { B } ^ { - 1 }$.
\hfill \mbox{\textit{OCR FP1 2006 Q6 [7]}}