FP1 June 2011 Q3
3.
(a) Given that \[\mathbf{A} = \begin{pmatrix} 1 & \sqrt{2} \\ \sqrt{2} & -1 \end{pmatrix}\]
(i) find \(\mathbf{A}^2\),
(ii) describe fully the geometrical transformation represented by \(\mathbf{A}^2\).
(4)(b) Given that \[\mathbf{B} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\] describe fully the geometrical transformation represented by \(\mathbf{B}\). (2)
(c) Given that \[\mathbf{C} = \begin{pmatrix} k+1 & 12 \\ k & 9 \end{pmatrix}\] where \(k\) is a constant, find the value of \(k\) for which the matrix \(\mathbf{C}\) is singular. (3)
| Scheme | Marks |
|---|---|
| \(\mathbf{A} = \begin{pmatrix} 1 & \sqrt{2} \\ \sqrt{2} & -1 \end{pmatrix}\) | |
| (i) \(\mathbf{A}^2 = \begin{pmatrix} 1 & \sqrt{2} \\ \sqrt{2} & -1 \end{pmatrix}\begin{pmatrix} 1 & \sqrt{2} \\ \sqrt{2} & -1 \end{pmatrix}\) | |
| \(= \begin{pmatrix} 1 + 2 & \sqrt{2} - \sqrt{2} \\ \sqrt{2} - \sqrt{2} & 2 + 1 \end{pmatrix}\) A correct method to multiply out two matrices. Can be implied by two out of four correct elements. | M1 |
| \(= \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\) Correct answer | A1 |
| (2) | |
| (ii) Enlargement; scale factor 3, centre \((0,\ 0)\). Enlargement; scale factor 3, centre (0, 0) | B1; B1 |
| (2) |
Notes
Allow ‘from’ or ‘about’ for centre and ‘O’ or ‘origin’ for (0, 0)
| Scheme | Marks |
|---|---|
| \(\mathbf{B} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) | |
| Reflection; in the line \(y = -x\). Reflection; \(y = -x\) | B1; B1 |
| (2) |
Notes
Allow ‘in the axis’ ‘about the line’ \(y = -x\) etc.
The question does not specify a single transformation so we would need to accept any combinations that are correct e.g. Anticlockwise rotation of \(90^\circ\) about the origin followed by a reflection in the \(x\)-axis is acceptable. In cases like these, the combination has to be completely correct and scored as B2 (no part marks). If in doubt consult your Team Leader.
| Scheme | Marks |
|---|---|
| \(\mathbf{C} = \begin{pmatrix} k+1 & 12 \\ k & 9 \end{pmatrix}\), \(k\) is a constant. | |
| \(\mathbf{C}\) is singular \(\Rightarrow \det\mathbf{C} = 0\). (Can be implied) \(\det\mathbf{C} = 0\) | B1 |
| \(9(k+1) - 12k\ (= 0)\) Applies \(9(k+1) - 12k\) | M1 |
| \(9k + 9 = 12k\) | |
| \(9 = 3k\) | |
| \(k = 3\) \(k = 3\) | A1 |
| (3) | |
| (9 marks) |
Notes
Special Case \(\dfrac{1}{9(k+1) - 12k} = 0\) B1(implied)M0A0
\(k = 3\) with no working can score full marks