Algebra of matrices

At a Glance

• Frequency: 5 sub-parts across 3 of 14 years (2015, 2016, 2021)
• Priority tier: T3
• Marks (count): 10 (2), 12 (1), 4 (1), 7 (1)
• Average solve time: ~5 min
• Difficulty mix: easy 3, medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

All five appearances are short, marks-efficient questions in Section A (compulsory or part of Section A optional). The two matrix-power questions reward knowing two algebraic patterns — nilpotency and cyclicity — that let you collapse a high power to something trivial without diagonalising. The trace identity question is a 3-minute proof that shows up as a stepping stone to the “commutator ≠ identity” result. None of these require deep theory; they reward systematic computation and pattern recognition.

Minimum Theory

Matrix powers. There are two key patterns: (i) Nilpotent: $A^k = O$ for some $k \ge 1$ (the index). Once you establish $A^m = O$, every higher power is also $O$. (ii) Cyclic of order $n$: $A^n = I$. Powers cycle with period $n$, so $A^{kn+r} = A^r$.

Nilpotency test. Compute $A^2$, then $A^3 = A \cdot A^2$, etc., until you reach $O$. If $A$ is $n \times n$ and nilpotent, $A^n = O$ (Cayley–Hamilton). If you suspect nilpotency: check $\operatorname{tr}(A) = 0$, $\det A = 0$, $\operatorname{tr}(A^2) = 0$. These are necessary conditions only; the actual index requires computation.

Trace formula. For $n \times n$ matrices $A$ and $B$: $\operatorname{tr}(AB) = \sum_{i=1}^n \sum_{k=1}^n a_{ik} b_{ki} = \operatorname{tr}(BA).$ The key consequence: the commutator $[A,B] = AB - BA$ is always traceless ($\operatorname{tr}(AB - BA) = 0$), so $AB - BA \ne cI$ for any nonzero scalar $c$.

Determinant limit (multilinearity). $\det$ is linear in each row. If row 1 of $\Delta(x)$ expands as row 1(0) + $x \cdot$(something) + $O(x^2)$, then: $\Delta(x) = \det(\text{row 1(0)}, R_2, R_3) + x \cdot \det(\text{something}, R_2, R_3) + O(x^2).$ If row 1(0) = $R_2$ (identical to row 2), the first determinant is 0; if “something” = $R_3$, the second is also 0 — giving $\Delta(x) = O(x^2)$ and $\lim_{x\to0}\Delta(x)/x = 0$.

Question Archetypes

matrix-power (2 questions; 2015, 2016)

Recognition Cues

• “Find $A^{30}$” (or other large power) for a given $3\times3$ matrix.
• The matrix is either nilpotent ($A^k = O$ for small $k$) or shows a parity pattern in powers.

Solution Template

1. Compute $A^2$, $A^3$. Look for $A^k = O$ (nilpotent) or $A^k = I$ (cyclic) or a stable block structure.
2. If nilpotent: state the index $k$ explicitly; conclude $A^n = O$ for all $n \ge k$.
3. If parity pattern: tabulate small powers, identify the even/odd formula, prove by induction (one step).
4. Apply to the target $n$.

Worked Example(s)

2016 Paper 1, 2016-P1-Q1a-ii (4 marks)

If $A = \begin{pmatrix}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{pmatrix}$, find $A^{14} + 3A - 2I$.

Nilpotency check. $\operatorname{tr}(A) = 1+2-3 = 0$. Compute $A^2$: numerically gives a nonzero matrix. Compute $A^3 = A \cdot A^2 = O$.

So $A$ is nilpotent of index 3. For $n \ge 3$, $A^n = O$. In particular, $A^{14} = O$.

$A^{14} + 3A - 2I = O + 3A - 2I = 3A - 2I = \begin{pmatrix}1 & 3 & 9 \\ 15 & 4 & 18 \\ -6 & -3 & -11\end{pmatrix}.$

2015 Paper 1, 2015-P1-Q2a (12 marks)

Find $A^{30}$ for $A = \begin{pmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}$.

Compute small powers: $A^2 = \begin{pmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{pmatrix}, \quad A^3 = \begin{pmatrix}1 & 0 & 0 \\ 2 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}, \quad A^4 = \begin{pmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1\end{pmatrix}.$

Pattern (by parity). For even $n = 2k$: $A^{2k} = \begin{pmatrix}1 & 0 & 0 \\ k & 1 & 0 \\ k & 0 & 1\end{pmatrix}.$ Check: $A^2$ ($k=1$) ✓; $A^4$ ($k=2$) ✓.

Inductive step. If the formula holds for $n=2k$, then $A^{2k+2} = A^{2k} \cdot A^2$: row 2 gives $k+1$ in position (2,1), row 3 gives $k+1$ in position (3,1). ✓

Apply $n=30$, $k=15$: $\boxed{A^{30} = \begin{pmatrix}1 & 0 & 0 \\ 15 & 1 & 0 \\ 15 & 0 & 1\end{pmatrix}.}$

Common Traps

• Nilpotency index is 3, not 2 in the 2016 example. You must compute $A^3$, not just $A^2$; concluding $A^2 = O$ is wrong and loses the whole answer.
• $A$ is not diagonalisable in the 2015 example (the eigenvalue $1$ has algebraic multiplicity 2 but only a 1-dimensional eigenspace). Attempting diagonalisation will fail; the parity-pattern approach is the right route.
• $\operatorname{tr}(A) = 0$ is a necessary condition for nilpotency but not sufficient. Always verify by computing powers.

matrix-identity-verification (1 question; 2021)

Recognition Cues

• “Show $A^2 = A^{-1}$” — equivalent to showing $A^3 = I$.
• “Show $A^k = I$” for some specific $k$.

Solution Template

1. Restate as a power identity: $A^2 = A^{-1} \Leftrightarrow A^3 = I$.
2. Compute $A^2$, then $A^3 = A \cdot A^2$. Verify $A^3 = I$ directly.
3. Conclude: $A \cdot A^2 = I \Rightarrow A^2 = A^{-1}$.

Worked Example

2021 Paper 1, 2021-P1-Q1a (10 marks)

Show $A^2 = A^{-1}$ (without computing $A^{-1}$) for $A = \begin{pmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{pmatrix}$.

$A^2 = A \cdot A$; row-by-row multiplication gives $A^2 = \begin{pmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{pmatrix}$.

Then $A^3 = A \cdot A^2$:

• Row 1: $(1)(-)\cdot\text{cols} = (1,0,0)$.
• Row 2: $(2)(-)\cdot\text{cols} = (0,1,0)$.
• Row 3: $(1)(-)\cdot\text{cols} = (0,0,1)$.

$A^3 = I$. Therefore $A \cdot A^2 = I$, so $A^2 = A^{-1}$. $\square$

Common Traps

• Compute $A^3 = A \cdot A^2$, not $(A^2)^2$. The latter gives $A^4$, which is not what you need.
• Verify $\det A \ne 0$ (so $A^{-1}$ exists). Here $\det A = 1$.

determinant-limit (1 question; 2021)

Recognition Cues

• $\lim_{x\to 0} \Delta(x)/x$ where $\Delta(x)$ is a determinant whose rows include terms like $f(x + k\alpha)$.
• At $x=0$, two rows become identical $\Rightarrow \Delta(0) = 0$, so the limit is a $0/0$ form.

Solution Template

1. Identify the $x$-dependent row. Write $f(x+k\alpha) = f(k\alpha) + x f'(k\alpha) + O(x^2)$ by Taylor expansion.
2. Decompose using multilinearity: $\Delta(x) = \Delta_0 + x \Delta_1 + O(x^2)$, where $\Delta_0$ and $\Delta_1$ are determinants with the expanded row replaced by its leading/next terms.
3. Check for identical rows in $\Delta_0$ and $\Delta_1$. If both are 0, then $\Delta(x) = O(x^2)$ and the limit is 0.

Worked Example

2021 Paper 1, 2021-P1-Q1c (10 marks)

$\Delta(x) = \begin{vmatrix}f(x+\alpha) & f(x+2\alpha) & f(x+3\alpha) \\ f(\alpha) & f(2\alpha) & f(3\alpha) \\ f'(\alpha) & f'(2\alpha) & f'(3\alpha)\end{vmatrix}$. Find $\lim_{x\to0}\Delta(x)/x$.

Taylor expansion of row 1: $f(x+k\alpha) = f(k\alpha) + x f'(k\alpha) + O(x^2)$.

So row 1 $=$ row 2 $+ x \cdot$ row 3 $+ O(x^2)$.

By multilinearity in row 1: $\Delta(x) = \underbrace{\det(\text{row 2}, \text{row 2}, \text{row 3})}_{=0} + x \cdot \underbrace{\det(\text{row 3}, \text{row 2}, \text{row 3})}_{=0} + O(x^2) = O(x^2).$

Both determinants are 0 because they each have two identical rows. Therefore $\Delta(x)/x \to 0$.

$\boxed{\lim_{x\to0}\frac{\Delta(x)}{x} = 0.}$

Common Traps

• The $O(x)$ term in $\Delta(x)$ has rows (row 3, row 2, row 3) — rows 1 and 3 are identical, giving a second zero. Both cancellations must be observed.

trace-property (1 question; 2021)

Recognition Cues

• “Show $\operatorname{tr}(AB) = \operatorname{tr}(BA)$; hence $AB - BA \ne I$.”
• The second part always follows from the first by a one-line contradiction.

Solution Template

1. Prove $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ via $(AB)_{ii} = \sum_k a_{ik}b_{ki}$; swap dummy indices.
2. Apply: $\operatorname{tr}(AB - BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = 0$, but $\operatorname{tr}(I_n) = n \ne 0$. Contradiction.

Worked Example

2021 Paper 1, 2021-P1-Q3c-ii (7 marks)

Show $\operatorname{tr}(AB) = \operatorname{tr}(BA)$; hence $AB - BA \ne I_2$.

Step 1. $(AB)_{ii} = \sum_k a_{ik} b_{ki}$, so $\operatorname{tr}(AB) = \sum_i\sum_k a_{ik} b_{ki}$. Swap indices $i \leftrightarrow k$: $= \sum_k\sum_i a_{ki}b_{ik} = \operatorname{tr}(BA)$. ✓

Step 2. If $AB - BA = I_2$, then $\operatorname{tr}(AB - BA) = \operatorname{tr}(I_2) = 2$. But $\operatorname{tr}(AB - BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = 0$. Contradiction. Hence $AB - BA \ne I_2$.

Common Traps

• The proof works for any $n \times n$ matrices, not just $2 \times 2$. The conclusion generalises: no finite-dimensional commutator can equal a nonzero multiple of the identity.
• Swap the dummy variables explicitly in the proof — don’t just say “by symmetry.”

Marks-Aware Writing

4-mark nilpotent question: Two lines suffice: compute $A^3 = O$ (show the computation, or state it is easily verified), then write $A^{14} = (A^3)^4 A^2 = O$, giving the final expression.

12-mark pattern question: Tabulate $A^2$, $A^3$, $A^4$ explicitly; state the parity formula; prove it by one inductive step; apply. Show each matrix product.

7-mark trace question: Two clearly labelled parts. Part 1: the index-swap computation (3 lines). Part 2: the contradiction argument (2 lines). Total 5–6 lines of clean algebra.

Practice Set

• 2018-P1-Q1a (10 m) — matrix trace/power variant;