Proving Statements Involving Matrices
A formula for Mⁿ is just another statement waiting for induction — multiply the assumed form by M once more, and check every entry lines up.
Induction with matrix powers
To prove $M^n=f(n)$ for a given matrix formula $f(n)$: verify the basis case by direct substitution; assume $M^k=f(k)$; then compute $M^{k+1}=M^k\cdot M$ using the assumed form, multiply out, and show the result equals $f(k+1)$.
Always write $M^{k+1}=M^k\times M$ (multiplying the assumed form by $M$ on the right), consistently in this order — although matrix multiplication is not commutative in general, using a fixed, standard order throughout avoids errors and matches the convention examiners expect.
Basis ($n=1$): $M^1=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, matching the formula with $n=1$.
Assumption: assume $M^k=\begin{pmatrix}1&k\\0&1\end{pmatrix}$ for some $k\geq1$.
Inductive step: $M^{k+1}=M^k\cdot M=\begin{pmatrix}1&k\\0&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}=\begin{pmatrix}1&k+1\\0&1\end{pmatrix}$, matching the formula with $n=k+1$.
Conclusion: since true for $n=1$, and true for $n=k+1$ whenever true for $n=k$, by induction $M^n=\begin{pmatrix}1&n\\0&1\end{pmatrix}$ for all positive integers $n$.
Basis ($n=1$): $M^1=\begin{pmatrix}2&0\\0&3\end{pmatrix}=\begin{pmatrix}2^1&0\\0&3^1\end{pmatrix}$, true.
Assumption: assume $M^k=\begin{pmatrix}2^k&0\\0&3^k\end{pmatrix}$ for some $k\geq1$.
Inductive step: $M^{k+1}=M^k\cdot M=\begin{pmatrix}2^k&0\\0&3^k\end{pmatrix}\begin{pmatrix}2&0\\0&3\end{pmatrix}=\begin{pmatrix}2^{k+1}&0\\0&3^{k+1}\end{pmatrix}$, matching the formula at $n=k+1$.
Conclusion: true for $n=1$; true for $n=k+1$ whenever true for $n=k$; so true for all positive integers $n$ by induction.
Induction with a non-diagonal, more intricate matrix
Basis ($n=1$): $M^1=\begin{pmatrix}3&-1\\4&-1\end{pmatrix}$. Formula at $n=1$: $\begin{pmatrix}3&-1\\4&-1\end{pmatrix}$. Matches.
Assumption: assume $M^k=\begin{pmatrix}2k+1&-k\\4k&1-2k\end{pmatrix}$ for some $k\geq1$.
Inductive step: $M^{k+1}=M^k\cdot M=\begin{pmatrix}2k+1&-k\\4k&1-2k\end{pmatrix}\begin{pmatrix}3&-1\\4&-1\end{pmatrix}$
Top-left: $(2k+1)(3)+(-k)(4)=6k+3-4k=2k+3=2(k+1)+1$ ✓
Top-right: $(2k+1)(-1)+(-k)(-1)=-2k-1+k=-k-1=-(k+1)$ ✓
Bottom-left: $4k(3)+(1-2k)(4)=12k+4-8k=4k+4=4(k+1)$ ✓
Bottom-right: $4k(-1)+(1-2k)(-1)=-4k-1+2k=-2k-1=1-2(k+1)$ ✓
All four entries match the formula at $n=k+1$.
Conclusion: true for $n=1$; true for $n=k+1$ whenever true for $n=k$; so true for all positive integers $n$ by induction.
Induction with determinants
Basis ($n=1$): $\det(M)=1(1)-2(0)=1$. True.
Assumption: assume $\det(M^k)=1$.
Inductive step: using the multiplicative property $\det(AB)=\det(A)\det(B)$: $\det(M^{k+1})=\det(M^k\cdot M)=\det(M^k)\det(M)=1\times1=1$.
Conclusion: true for $n=1$; true for $n=k+1$ whenever true for $n=k$; so $\det(M^n)=1$ for all positive integers $n$ by induction.
Forgetting to actually carry out the matrix multiplication $M^k\cdot M$ in full — writing down the "expected" answer without deriving it from the assumed form gives no genuine proof and will lose the method marks, even if the final matrix is correct.
Using a proven matrix formula: "hence" questions
Once a formula for $M^n$ has been established by induction, exam questions often ask you to use that result — for example, to find the value of $n$ that produces a given matrix, or to combine $M^n$ with another matrix. These "hence" parts don't need a fresh induction; they just require substituting into the formula you already proved and solving.
Comparing the top-right entry of the proven formula with the target matrix: $3n=45$, so $n=15$. (The other three entries, $1$, $0$, $1$, already match for every $n$, so no further condition is needed.)
Exam-style questions
Question 1 6 marks
Given $M=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, prove by induction that $M^n=\begin{pmatrix}1&0\\n&1\end{pmatrix}$ for all positive integers $n$.
Basis ($n=1$): $M^1=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, matches the formula. (B1)
Assumption: assume $M^k=\begin{pmatrix}1&0\\k&1\end{pmatrix}$. (M1)
Inductive step: $M^{k+1}=M^k M=\begin{pmatrix}1&0\\k&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}1&0\\k+1&1\end{pmatrix}$, matching $n=k+1$. (M1, A1, A1)
Conclusion: true for $n=1$; true for $n=k+1$ whenever true for $n=k$; so true for all positive integers $n$. (B1)
Question 2 8 marks
Given $M=\begin{pmatrix}5&-4\\4&-3\end{pmatrix}$, prove by induction that $M^n=\begin{pmatrix}4n+1&-4n\\4n&1-4n\end{pmatrix}$ for all positive integers $n$.
Basis ($n=1$): matches directly. (B1)
Assumption: assume $M^k=\begin{pmatrix}4k+1&-4k\\4k&1-4k\end{pmatrix}$. (M1)
Inductive step: $M^{k+1}=M^kM$; top-left: $(4k+1)(5)+(-4k)(4)=20k+5-16k=4k+5=4(k+1)+1$ ✓ (M1, A1)
Remaining three entries verified similarly, all matching the formula at $n=k+1$. (A1, A1, A1)
Conclusion: true for $n=1$; true for $n=k+1$ whenever true for $n=k$; so true for all positive integers $n$. (B1)
Question 3 5 marks
Given a $2\times2$ matrix $M$ with $\det(M)=2$, prove by induction that $\det(M^n)=2^n$ for all positive integers $n$.
Basis ($n=1$): $\det(M^1)=\det(M)=2=2^1$. (B1)
Assumption: assume $\det(M^k)=2^k$. (M1)
Inductive step: $\det(M^{k+1})=\det(M^kM)=\det(M^k)\det(M)=2^k\times2=2^{k+1}$ (M1, A1)
Conclusion: true for $n=1$; true for $n=k+1$ whenever true for $n=k$; so true for all positive integers $n$. (B1)