SPS SPS FM Pure 2022 February — Question 7 5 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2022
SessionFebruary
Marks5
TopicProof by induction
TypeProve matrix power formula
DifficultyStandard +0.8 This is a Further Maths proof by induction involving matrix multiplication. While the structure is standard (base case, inductive step, matrix multiplication), students must correctly multiply 3×3 matrices and manipulate expressions involving powers of 3. The algebraic manipulation in the top-right entries requires care but follows a predictable pattern. More demanding than typical A-level questions due to matrix multiplication complexity, but still a routine Further Maths induction proof.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar
The matrix \(\mathbf{M}\) is defined by \(\mathbf{M} = \begin{pmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) $$\mathbf{M}^n = \begin{pmatrix} 3^n & 3^n - 1 & -3^n + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ Prove by induction that \(\mathbf{M}^n = \begin{pmatrix} 3^n & 3^n - 1 & -3^n + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) for all integers \(n \geq 1\) [5 marks] 
The matrix $\mathbf{M}$ is defined by $\mathbf{M} = \begin{pmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$$\mathbf{M}^n = \begin{pmatrix} 3^n & 3^n - 1 & -3^n + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Prove by induction that $\mathbf{M}^n = \begin{pmatrix} 3^n & 3^n - 1 & -3^n + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ for all integers $n \geq 1$

[5 marks]

\hfill \mbox{\textit{SPS SPS FM Pure 2022 Q7 [5]}}
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