Standard +0.8 This question requires computing two 3×3 matrix products (AB and BA), setting them equal, and solving for the parameter a. While the matrix multiplication itself is routine, the problem involves substantial computation (18 entries to calculate), careful algebraic manipulation, and the conceptual understanding that matrix multiplication is generally non-commutative. This is more demanding than standard matrix multiplication exercises but doesn't require deep theoretical insight.
1 Two matrices, \(\mathbf { A }\) and \(\mathbf { B }\), are given by \(\mathbf { A } = \left( \begin{array} { r r r } 1 & - 2 & - 1 \\ 2 & - 3 & 1 \\ a & 1 & 1 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { r r r } - 6 & 3 & - 4 \\ - 1 & 6 & - 4 \\ 8 & - 8 & - 1 \end{array} \right)\) where \(a\) is a constant. Find the value of \(a\) for which \(\mathbf { A B } = \mathbf { B A }\).
Question 1:
1 | −12 −1 5
−1 −20 3 or
7−6a 3a−2 −4a−5
−4a −1 5
11−4a −20 3 seen
−8−a 7 −17
–12 = –4a or –1 = 11 – 4a or 7 – 6a = –8 – a or
3a – 2 = 7 or –4a – 5 = –17
a = 3 | M1
M1
A1
[3] | 1.1
1.1
2.2a | Either product AB or BA
calculated (but not if assigned
incorrectly).
Alternatively: equivalent correct
useful entries calculated for both
Finding matrix products both
ways and equating entries usefully | Condone 3 errors or omissions
This mark can be implied by sight
of a correct equation
This mark can be implied by sight
of a correct equation even if other
entries or equations are wrong.
Cannot be awarded if either AB or
BA has more than 3 errors
1 Two matrices, $\mathbf { A }$ and $\mathbf { B }$, are given by $\mathbf { A } = \left( \begin{array} { r r r } 1 & - 2 & - 1 \\ 2 & - 3 & 1 \\ a & 1 & 1 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r r } - 6 & 3 & - 4 \\ - 1 & 6 & - 4 \\ 8 & - 8 & - 1 \end{array} \right)$ where $a$ is a constant. Find the value of $a$ for which $\mathbf { A B } = \mathbf { B A }$.
\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q1 [3]}}