Standard +0.8 This is a standard proof by induction for matrix powers with a clear pattern. While it requires matrix multiplication and careful algebraic manipulation (especially the 5n term in the top-right entry), it follows a routine induction structure with no novel insights needed. Slightly above average due to being Further Maths content and requiring careful handling of the algebraic terms, but well within the standard repertoire of induction proofs.
7. Using mathematical induction, prove that
$$\left[ \begin{array} { l l }
2 & 5 \\
0 & 2
\end{array} \right] ^ { n } = \left[ \begin{array} { c c }
2 ^ { n } & 2 ^ { n - 1 } \times 5 n \\
0 & 2 ^ { n }
\end{array} \right]$$
for all positive integers \(n\).
If proposition is true for \(n = k\), it is also true for \(n = k + 1\). As it is true for \(n = 1\), by mathematical induction, it is true for all positive integers \(n\).
E1
Award for a perfect solution including the last line.
Total: [7]
When $n = 1$, LHS $= \begin{pmatrix} 2 & 5 \\ 0 & 2 \end{pmatrix}$ and RHS $= \begin{pmatrix} 2 & 5 \\ 0 & 2 \end{pmatrix}$ | B1 |
Therefore, proposition is valid for $n = 1$.
Assume result is true for $n = k$ i.e. $\begin{pmatrix} 2 & 5 \\ 0 & 2 \end{pmatrix}^k = \begin{pmatrix} 2^k & 2^{k-1} \times 5k \\ 0 & 2^k \end{pmatrix}$ | M1 |
Consider $n = k + 1$: $\begin{pmatrix} 2 & 5 \\ 0 & 2 \end{pmatrix}^{k+1} = \begin{pmatrix} 2 & 5 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2^k & 2^{k-1} \times 5k \\ 0 & 2^k \end{pmatrix}$ | M1 |
$= \begin{pmatrix} 2 \times 2^k & (2 \times 2^{k-1} \times 5k) + (5 \times 2^k) \\ 0 & 2 \times 2^k \end{pmatrix}$ | A1 | Or $\begin{pmatrix} 2^k & 2^{k-1} \times 5k \\ 0 & 2^k \end{pmatrix}\begin{pmatrix} 2 & 5 \end{pmatrix}$
Top right entry: $(2^k \times 5k) + (5 \times 2^k) = 2^k(5k + 5) = 2^k \times 5(k + 1)$ | A1 |
Therefore, $\begin{pmatrix} 2 & 5 \\ 0 & 2 \end{pmatrix}^{k+1} = \begin{pmatrix} 2^{k+1} & 2^k \times 5(k + 1) \\ 0 & 2^{k+1} \end{pmatrix}$ | A1 | Remaining 3 entries correct
If proposition is true for $n = k$, it is also true for $n = k + 1$. As it is true for $n = 1$, by mathematical induction, it is true for all positive integers $n$. | E1 | Award for a perfect solution including the last line.
**Total: [7]**
7. Using mathematical induction, prove that
$$\left[ \begin{array} { l l }
2 & 5 \\
0 & 2
\end{array} \right] ^ { n } = \left[ \begin{array} { c c }
2 ^ { n } & 2 ^ { n - 1 } \times 5 n \\
0 & 2 ^ { n }
\end{array} \right]$$
for all positive integers $n$.\\
\hfill \mbox{\textit{WJEC Further Unit 1 2023 Q7 [7]}}