Standard +0.8 This is a Further Maths question requiring students to set AB = BA, expand both products, equate corresponding entries to form a system of equations, and solve for p. While the matrix multiplication itself is routine, the problem-solving aspect (recognizing what 'commutative' means and systematically working through the algebra) and the need to fully justify the answer elevates this above a standard question. It's moderately challenging but follows a clear method once the approach is identified.
The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are defined as follows.
$$\mathbf{A} = \begin{bmatrix} p - 2 & p - 1 \\ 0 & 1 \end{bmatrix} \quad\quad \mathbf{B} = \begin{bmatrix} 1 & 2p - 1 \\ 0 & 4 - p \end{bmatrix}$$
Find the values of \(p\) such that \(\mathbf{A}\) and \(\mathbf{B}\) are commutative under matrix multiplication.
Fully justify your answer.
[4 marks]
Question 7:
7 | Calculates AB or BA with at
least three correct elements. | 1.1a | M1 | p–2 p–11 2p–1
AB=
0 1 0 4– p
p–2 (p−2)(2p−1)+(p−1)(4−p)
=
0 4– p
p–2 p2–2
=
0 4– p
1 2p–1p–2 p–1
BA=
0 4– p 0 1
p–2 p−1+2p−1
=
0 4– p
p–2 3p−2
=
0 4– p
A and B are commutative, hence
AB = BA
Thus
p2 – 2 = 3p – 2
p(p – 3) = 0
p = 0, 3
Calculates AB and BA with at
least seven correct elements. | 1.2 | M1
Uses AB = BA to form a
quadratic equation in p and
obtains at least one solution | 1.1a | M1
Completes a reasoned
argument to obtain p = 0, 3
Must have stated that since A
and B are commutative (under
matrix multiplication,)
AB = BA | 2.1 | R1
Question total | 4
Q | Marking
instructions | AO | Marks | Typical solution
The matrices $\mathbf{A}$ and $\mathbf{B}$ are defined as follows.
$$\mathbf{A} = \begin{bmatrix} p - 2 & p - 1 \\ 0 & 1 \end{bmatrix} \quad\quad \mathbf{B} = \begin{bmatrix} 1 & 2p - 1 \\ 0 & 4 - p \end{bmatrix}$$
Find the values of $p$ such that $\mathbf{A}$ and $\mathbf{B}$ are commutative under matrix multiplication.
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 2 2024 Q7 [4]}}