Standard +0.8 This is a proof by induction involving matrix algebra, requiring understanding of matrix multiplication properties and inverse operations. While the algebraic manipulation is relatively straightforward once the inductive step is set up (using associativity and A^{-1}A = I), it requires careful symbolic reasoning with non-commutative objects and proper induction structure. It's above average difficulty but not exceptionally hard for Further Maths students who have practiced matrix proofs.
11 Prove by induction that, for all integers \(n \geq 1\),
$$\left( \mathbf { A B A } ^ { - 1 } \right) ^ { n } = \mathbf { A B } ^ { n } \mathbf { A } ^ { - 1 }$$
where \(\mathbf { A }\) and \(\mathbf { B }\) are square matrices of equal dimensions, and \(\mathbf { A }\) is non-singular.
Completes rigorous working to show \((ABA^{-1})^{k+1} = AB^{k+1}A^{-1}\); condone \(A^{-1}A\) removed without reference to \(\mathbf{I}\)
Therefore by induction \((ABA^{-1})^n = AB^nA^{-1}\) is true for all integers \(n \geq 1\)
R1
Concludes reasoned argument stating true for \(n=1\), assumption implies \(n=k+1\), hence true for all \(n\geq 1\); condone \(A^{-1}A\) removed without reference to \(\mathbf{I}\)
## Question 11:
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $n=1$: $(ABA^{-1})^1 = ABA^{-1} = AB^1A^{-1}$; true for $n=1$ | B1 | Shows $(ABA^{-1})^n = AB^nA^{-1}$ is true for $n=1$ |
| Assumes $(ABA^{-1})^k = AB^kA^{-1}$ and multiplies by $ABA^{-1}$ | M1 | |
| $(ABA^{-1})^k \cdot ABA^{-1} = AB^kA^{-1}ABA^{-1} \Rightarrow AB^k\mathbf{I}BA^{-1} \Rightarrow AB^kBA^{-1} \Rightarrow AB^{k+1}A^{-1}$ | A1 | Completes rigorous working to show $(ABA^{-1})^{k+1} = AB^{k+1}A^{-1}$; condone $A^{-1}A$ removed without reference to $\mathbf{I}$ |
| Therefore by induction $(ABA^{-1})^n = AB^nA^{-1}$ is true for all integers $n \geq 1$ | R1 | Concludes reasoned argument stating true for $n=1$, assumption implies $n=k+1$, hence true for all $n\geq 1$; condone $A^{-1}A$ removed without reference to $\mathbf{I}$ |
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11 Prove by induction that, for all integers $n \geq 1$,
$$\left( \mathbf { A B A } ^ { - 1 } \right) ^ { n } = \mathbf { A B } ^ { n } \mathbf { A } ^ { - 1 }$$
where $\mathbf { A }$ and $\mathbf { B }$ are square matrices of equal dimensions, and $\mathbf { A }$ is non-singular.
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q11 [4]}}