← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q2b-ii — Step-by-Step Solution

10 marks · Section A

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

Question

Verify Cayley–Hamilton theorem for the matrix A=[1423]A=\begin{bmatrix}1 & 4\\ 2 & 3\end{bmatrix} and hence find its inverse. Also, find the matrix represented by A5−4A4−7A3+11A2−A−10IA^5-4A^4-7A^3+11A^2-A-10I.

Technique

Cayley–Hamilton reduces all higher powers AkA^k (k≥2k\ge 2) to linear combinations of AA and II.

Solution

Step 1 — Characteristic polynomial and Cayley–Hamilton

det⁡(A−λI)=(1−λ)(3−λ)−(4)(2)=λ2−4λ−5.\det(A-\lambda I)=(1-\lambda)(3-\lambda)-(4)(2)=\lambda^{2}-4\lambda-5.

Cayley–Hamilton: A2−4A−5I=0A^{2}-4A-5I=0.

Verify directly:

A2=(1423)(1423)=(916817).A^{2}=\begin{pmatrix}1 & 4\\ 2 & 3\end{pmatrix}\begin{pmatrix}1 & 4\\ 2 & 3\end{pmatrix}=\begin{pmatrix}9 & 16\\ 8 & 17\end{pmatrix}. 4A+5I=(416812)+(5005)=(916817)=A2  ✓.4A+5I=\begin{pmatrix}4 & 16\\ 8 & 12\end{pmatrix}+\begin{pmatrix}5 & 0\\ 0 & 5\end{pmatrix}=\begin{pmatrix}9 & 16\\ 8 & 17\end{pmatrix}=A^{2}\;\checkmark.

Step 2 — Inverse via Cayley–Hamilton

From A2−4A−5I=0A^{2}-4A-5I=0: A(A−4I)=5IA(A-4I)=5I, hence

A−1=A−4I5=15(−342−1)=(−3/54/52/5−1/5).A^{-1}=\frac{A-4I}{5}=\frac{1}{5}\begin{pmatrix}-3 & 4\\ 2 & -1\end{pmatrix}=\begin{pmatrix}-3/5 & 4/5\\ 2/5 & -1/5\end{pmatrix}.

(Check via standard formula: det⁡A=3−8=−5\det A=3-8=-5, A−1=1−5(3−4−21)=15(−342−1)A^{-1}=\frac{1}{-5}\begin{pmatrix}3 & -4\\ -2 & 1\end{pmatrix}=\frac{1}{5}\begin{pmatrix}-3 & 4\\ 2 & -1\end{pmatrix} ✓.)

Step 3 — Reduce powers of AA using A2=4A+5IA^{2}=4A+5I

A3=A⋅A2=A(4A+5I)=4A2+5A=4(4A+5I)+5A=21A+20I.A^{3}=A\cdot A^{2}=A(4A+5I)=4A^{2}+5A=4(4A+5I)+5A=21A+20I. A4=A⋅A3=A(21A+20I)=21A2+20A=21(4A+5I)+20A=104A+105I.A^{4}=A\cdot A^{3}=A(21A+20I)=21A^{2}+20A=21(4A+5I)+20A=104A+105I. A5=A⋅A4=A(104A+105I)=104A2+105A=104(4A+5I)+105A=521A+520I.A^{5}=A\cdot A^{4}=A(104A+105I)=104A^{2}+105A=104(4A+5I)+105A=521A+520I.

Step 4 — Evaluate A5−4A4−7A3+11A2−A−10IA^{5}-4A^{4}-7A^{3}+11A^{2}-A-10I

Substitute each power as αA+βI\alpha A+\beta I:

TermαA\alpha A-partβI\beta I-part
A5A^{5}521A521A520I520I
−4A4-4A^{4}−416A-416A−420I-420I
−7A3-7A^{3}−147A-147A−140I-140I
11A211A^{2}44A44A55I55I
−A-A−A-A—
−10I-10I—−10I-10I

Sum:

Result: A+5IA+5I.

A+5I=(1423)+(5005)=(6428).A+5I=\begin{pmatrix}1 & 4\\ 2 & 3\end{pmatrix}+\begin{pmatrix}5 & 0\\ 0 & 5\end{pmatrix}=\begin{pmatrix}6 & 4\\ 2 & 8\end{pmatrix}.

Answer

  A5−4A4−7A3+11A2−A−10I=(6428).  \boxed{\;A^{5}-4A^{4}-7A^{3}+11A^{2}-A-10I=\begin{pmatrix}6 & 4\\ 2 & 8\end{pmatrix}.\;}
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