A2 June 2024 Paper 2 Q7
7 The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are defined as follows.
\[\mathbf{A} = \begin{bmatrix} p - 2 & p - 1 \\ 0 & 1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 & 2p - 1 \\ 0 & 4 - p \end{bmatrix}\]Find the values of \(p\) such that \(\mathbf{A}\) and \(\mathbf{B}\) are commutative under matrix multiplication.
Fully justify your answer. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Calculates \(\mathbf{AB}\) or \(\mathbf{BA}\) with at least three correct elements. | M1 | 1.1a |
| Calculates \(\mathbf{AB}\) and \(\mathbf{BA}\) with at least seven correct elements. | M1 | 1.2 |
| Uses \(\mathbf{AB} = \mathbf{BA}\) to form a quadratic equation in \(p\) and obtains at least one solution | M1 | 1.1a |
| Completes a reasoned argument to obtain \(p = 0, 3\) Must have stated that since \(\mathbf{A}\) and \(\mathbf{B}\) are commutative (under matrix multiplication,) \(\mathbf{AB} = \mathbf{BA}\) | R1 | 2.1 |
| (4 marks) |
Typical solution
\[\begin{aligned} \mathbf{AB} &= \begin{bmatrix} p - 2 & p - 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2p - 1 \\ 0 & 4 - p \end{bmatrix} \\ &= \begin{bmatrix} p - 2 & (p - 2)(2p - 1) + (p - 1)(4 - p) \\ 0 & 4 - p \end{bmatrix} \\ &= \begin{bmatrix} p - 2 & p^2 - 2 \\ 0 & 4 - p \end{bmatrix} \end{aligned}\]\[\begin{aligned} \mathbf{BA} &= \begin{bmatrix} 1 & 2p - 1 \\ 0 & 4 - p \end{bmatrix}\begin{bmatrix} p - 2 & p - 1 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} p - 2 & p - 1 + 2p - 1 \\ 0 & 4 - p \end{bmatrix} \\ &= \begin{bmatrix} p - 2 & 3p - 2 \\ 0 & 4 - p \end{bmatrix} \end{aligned}\]\(\mathbf{A}\) and \(\mathbf{B}\) are commutative, hence \(\mathbf{AB} = \mathbf{BA}\)
Thus
\[p^2 - 2 = 3p - 2\]\[p(p - 3) = 0\]\[p = 0, 3\]