← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q3a — Step-by-Step Solution

20 marks · Section A

Cayley-Hamilton theorem · Linear Algebra · asked 3× in 13 yrs · Read the full method →

Question

Let A=[100101010]A=\begin{bmatrix}1 & 0 & 0\\1 & 0 & 1\\0 & 1 & 0\end{bmatrix}.

(i) Verify the Cayley–Hamilton theorem for the matrix AA.

(ii) Show that An=An−2+A2−IA^n=A^{n-2}+A^2-I for n≥3n\ge 3, where II is the identity matrix of order 3. Hence, find A40A^{40}.

Technique

(i) Direct CH verification by computing A2,A3A^2,A^3 and checking the polynomial vanishes; (ii) recurrence from CH gives An−An−2=A2−IA^n-A^{n-2}=A^2-I which telescopes for even step counts.

Solution

Part (i) — Verify Cayley–Hamilton

Characteristic polynomial.

det⁡(A−λI)=det⁡[1−λ001−λ101−λ]=(1−λ)det⁡[−λ11−λ]=(1−λ)(λ2−1)=−(λ−1)2(λ+1).\det(A-\lambda I)=\det\begin{bmatrix}1-\lambda & 0 & 0\\1 & -\lambda & 1\\0 & 1 & -\lambda\end{bmatrix}=(1-\lambda)\det\begin{bmatrix}-\lambda & 1\\1 & -\lambda\end{bmatrix}=(1-\lambda)(\lambda^2-1)=-(\lambda-1)^2(\lambda+1).

The characteristic equation is λ3−λ2−λ+1=0\lambda^3-\lambda^2-\lambda+1=0.

Compute A2,A3A^2,A^3.

A2=A⋅A=[100110101].A^2=A\cdot A=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\1 & 0 & 1\end{bmatrix}.

(Calculation: row 1 of AA times AA = (1,0,0)(1,0,0); row 2 of AA = (1,0,1)(1,0,1) — multiply: col 1 of AA = (1,1,0)(1,1,0) → 1+0+0=11+0+0=1; col 2 = (0,0,1)(0,0,1) → 0+0+1=10+0+1=1; col 3 = (0,1,0)(0,1,0) → 0+0+0=00+0+0=0. So row 2 = (1,1,0)(1,1,0). Row 3 of AA = (0,1,0)(0,1,0) → row 2 of AA = (1,0,1)(1,0,1).)

A3=A⋅A2=[100201110].A^3=A\cdot A^2=\begin{bmatrix}1 & 0 & 0\\2 & 0 & 1\\1 & 1 & 0\end{bmatrix}.

Cayley–Hamilton states A3−A2−A+I=0A^3-A^2-A+I=0. Compute:

A3−A2−A+I=[1−1−1+10−0−0+00−0−0+02−1−1+00−1−0+01−0−1+01−1−0+01−0−1+00−1−0+1]=[0000−10000].A^3-A^2-A+I=\begin{bmatrix}1-1-1+1 & 0-0-0+0 & 0-0-0+0\\2-1-1+0 & 0-1-0+0 & 1-0-1+0\\1-1-0+0 & 1-0-1+0 & 0-1-0+1\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\0 & -1 & 0\\0 & 0 & 0\end{bmatrix}.

Wait — the (2,2)(2,2) entry is −1-1, not 00. Let me recompute A2A^2 more carefully.

A=[100101010]A=\begin{bmatrix}1 & 0 & 0\\1 & 0 & 1\\0 & 1 & 0\end{bmatrix}. A2=A⋅AA^2=A\cdot A:

Row ii, column jj: ∑kAikAkj\sum_k A_{ik}A_{kj}.

A2=[100110101].A^2=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\1 & 0 & 1\end{bmatrix}.

A3=A⋅A2A^3=A\cdot A^2:

A3=[100201110].A^3=\begin{bmatrix}1 & 0 & 0\\2 & 0 & 1\\1 & 1 & 0\end{bmatrix}.

A3−A2−A+IA^3-A^2-A+I:

All zero ✓. Cayley–Hamilton verified.

Part (ii) — Recurrence and A40A^{40}

From CH: A3=A2+A−IA^3=A^2+A-I.

Claim: An=An−2+A2−IA^n=A^{n-2}+A^2-I for all n≥3n\ge 3.

Proof by induction. Base n=3n=3: A3=A+A2−I=A2+A−IA^3=A+A^2-I=A^2+A-I ✓ (matches CH).

Inductive step. Assume Ak=Ak−2+A2−IA^k=A^{k-2}+A^2-I. Then

Ak+1=A⋅Ak=A(Ak−2+A2−I)=Ak−1+A3−A=Ak−1+(A2+A−I)−A=Ak−1+A2−I.A^{k+1}=A\cdot A^k=A(A^{k-2}+A^2-I)=A^{k-1}+A^3-A=A^{k-1}+(A^2+A-I)-A=A^{k-1}+A^2-I.

This is the claim with n=k+1n=k+1. ✓

Compute A40A^{40}.

Using An−An−2=A2−IA^n-A^{n-2}=A^2-I repeatedly:

A40−A38=A2−I,A38−A36=A2−I,…,A4−A2=A2−I.A^{40}-A^{38}=A^2-I,\quad A^{38}-A^{36}=A^2-I,\quad\ldots,\quad A^4-A^2=A^2-I.

There are 1919 such equations (steps from A2A^2 up to A40A^{40} in increments of 2). Telescope:

A40−A2=19(A2−I)⇒A40=A2+19A2−19I=20A2−19I.A^{40}-A^2=19(A^2-I)\Rightarrow A^{40}=A^2+19A^2-19I=20A^2-19I.

Plug in A2A^2:

A40=20[100110101]−19[100010001]=[10020102001].A^{40}=20\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\1 & 0 & 1\end{bmatrix}-19\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\20 & 1 & 0\\20 & 0 & 1\end{bmatrix}.

Answer

  A40=[10020102001].  \boxed{\;A^{40}=\begin{bmatrix}1 & 0 & 0\\20 & 1 & 0\\20 & 0 & 1\end{bmatrix}.\;}
We post more of this — worked solutions, CSAT trap breakdowns, guide chapters — a few times a week on Telegram. Free, no sign-in. Join

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.