A2 October 2021 Paper 2 Q1
1 Two matrices, \(\mathbf{A}\) and \(\mathbf{B}\), are given by \(\mathbf{A} = \begin{pmatrix} 1 & -2 & -1 \\ 2 & -3 & 1 \\ a & 1 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} -6 & 3 & -4 \\ -1 & 6 & -4 \\ 8 & -8 & -1 \end{pmatrix}\) where \(a\) is a constant.
Find the value of \(a\) for which \(\mathbf{AB} = \mathbf{BA}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix} -12 & -1 & 5 \\ -1 & -20 & 3 \\ 7 - 6a & 3a - 2 & -4a - 5 \end{pmatrix}\) or \(\mathbf{BA} = \begin{pmatrix} -4a & -1 & 5 \\ 11 - 4a & -20 & 3 \\ -8 - a & 7 & -17 \end{pmatrix}\) seen | M1 | 1.1 |
| \(-12 = -4a\) or \(-1 = 11 - 4a\) or \(7 - 6a = -8 - a\) or \(3a - 2 = 7\) or \(-4a - 5 = -17\) | M1 | 1.1 |
| \(a = 3\) | A1 | 2.2a |
| [3] |
Notes
M1: Either product AB or BA calculated (but not if assigned incorrectly). Alternatively: equivalent correct useful entries calculated for both
Condone 3 errors or omissions
This mark can be implied by sight of a correct equation
M1: Finding matrix products both ways and equating entries usefully
This mark can be implied by sight of a correct equation even if other entries or equations are wrong.
A1: Cannot be awarded if either AB or BA has more than 3 errors