SPS SPS ASFM 2020 May — Question 6 6 marks

Exam BoardSPS
ModuleSPS ASFM (SPS ASFM)
Year2020
SessionMay
Marks6
TopicProof by induction
TypeProve matrix power formula
DifficultyChallenging +1.2 This is a proof by induction on matrix powers with a non-diagonal matrix. While it requires understanding of matrix multiplication and induction structure, the pattern is given explicitly, the base case is straightforward, and the inductive step involves routine matrix multiplication with simple algebraic manipulation. It's moderately harder than average due to the proof requirement and matrix context, but follows a standard template without requiring novel insight.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar

In this question you must show detailed reasoning. M is the matrix \(\begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix}\). Prove that \(\mathbf{M}^n = \begin{pmatrix} 1 & 3(2^{n+1} - 2) \\ 0 & 2^n \end{pmatrix}\), for any positive integer \(n\). [6]

\textbf{In this question you must show detailed reasoning.}

M is the matrix $\begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix}$.

Prove that $\mathbf{M}^n = \begin{pmatrix} 1 & 3(2^{n+1} - 2) \\ 0 & 2^n \end{pmatrix}$, for any positive integer $n$. [6]

\hfill \mbox{\textit{SPS SPS ASFM 2020 Q6 [6]}}