FP1 June 2013 (R) Q2
2.
(i) \[\mathbf{A} = \begin{pmatrix} 2k + 1 & k \\ -3 & -5 \end{pmatrix}, \quad \text{where } k \text{ is a constant}\]
Given that \[\mathbf{B} = \mathbf{A} + 3\mathbf{I}\] where \(\mathbf{I}\) is the \(2 \times 2\) identity matrix, find
(a) \(\mathbf{B}\) in terms of \(k\), (2)
(b) the value of \(k\) for which \(\mathbf{B}\) is singular. (2)
(ii) Given that \[\mathbf{C} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 2 & -1 & 5 \end{pmatrix}\] and \[\mathbf{E} = \mathbf{CD}\] find \(\mathbf{E}\). (2)
| Scheme | Marks |
|---|---|
| \(\mathbf{B} = \mathbf{A} + 3\mathbf{I} = \begin{pmatrix} 2k + 1 & k \\ -3 & -5 \end{pmatrix} + 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) For applying \(\mathbf{A} + 3\mathbf{I}\). Can be implied by three out of four correct elements in candidate’s final answer. Solution must come from addition. | M1 |
| \(= \begin{pmatrix} 2k + 4 & k \\ -3 & -2 \end{pmatrix}\) Correct answer. | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{B}\) is singular \(\Rightarrow \det\mathbf{B} = 0\). | |
| \(-2(2k + 4) - (-3k) = 0\) Applies “\(ad - bc\)” to \(\mathbf{B}\) and equates to 0 | M1 |
| \(-4k - 8 + 3k = 0\) | |
| \(k = -8\) \(k = -8\) | A1cao |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{C} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}, \mathbf{D} = \begin{pmatrix} 2 & -1 & 5 \end{pmatrix}, \mathbf{E} = \mathbf{CD}\) | |
| \(\mathbf{E} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}\begin{pmatrix} 2 & -1 & 5 \end{pmatrix} = \begin{pmatrix} 4 & -2 & 10 \\ -6 & 3 & -15 \\ 8 & -4 & 20 \end{pmatrix}\) Candidate writes down a \(3 \times 3\) matrix. Correct answer. | M1 A1 |
| (2) | |
| (6 marks) |