FP1 June 2014 Q4
4.
(i) Given that \[\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & -1 \\ 4 & 5 \end{pmatrix} \text{ and } \mathbf{B} = \begin{pmatrix} 2 & -1 & 4 \\ 1 & 3 & 1 \end{pmatrix},\]
(a) find \(\mathbf{AB}\).
(b) Explain why \(\mathbf{AB} \neq \mathbf{BA}\). (4)
(ii) Given that \[\mathbf{C} = \begin{pmatrix} 2k & -2 \\ 3 & k \end{pmatrix}, \text{ where } k \text{ is a real number}\] find \(\mathbf{C}^{-1}\), giving your answer in terms of \(k\). (3)
| Scheme | Marks |
|---|---|
| \(\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & -1 \\ 4 & 5 \end{pmatrix}, \mathbf{B} = \begin{pmatrix} 2 & -1 & 4 \\ 1 & 3 & 1 \end{pmatrix}\) | |
| \(\begin{pmatrix} 1 & 2 \\ 3 & -1 \\ 4 & 5 \end{pmatrix}\begin{pmatrix} 2 & -1 & 4 \\ 1 & 3 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 5 & 6 \\ 5 & -6 & 11 \\ 13 & 11 & 21 \end{pmatrix}\) M1: 3x3 matrix with a number or numerical expression for each element A2: cao (\(-1\) each error) Only 1 error award A1A0 | M1A2 |
| Scheme | Marks |
|---|---|
| \(\mathbf{BA} = \begin{pmatrix} 2 & -1 & 4 \\ 1 & 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & -1 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} 15 & 25 \\ 14 & 4 \end{pmatrix}\) Allow any convincing argument. E.g.s BA is a 2x2 matrix (so \(\mathbf{AB} \neq \mathbf{BA}\)) or dimensionally different. Attempt to evaluate product not required. NB ‘Not commutative’ only is B0 | B1 |
| (4) |
| Scheme | Marks |
|---|---|
| \((\det\mathbf{C} =)\, 2k \times k - 3 \times (-2)\) Correct attempt at determinant | M1 |
| \(\mathbf{C}^{-1} = \dfrac{1}{2k^2 + 6}\begin{pmatrix} k & 2 \\ -3 & 2k \end{pmatrix}\) M1: \(\dfrac{1}{\text{their } \det\mathbf{C}}\begin{pmatrix} k & 2 \\ -3 & 2k \end{pmatrix}\) A1: cao oe | M1A1 |
| (3) | |
| (7 marks) |