← 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 14 yrs · Read the full method →

Question

Verify Cayley–Hamilton theorem for the matrix $A=\begin{bmatrix}1 & 4\\ 2 & 3\end{bmatrix}$ and hence find its inverse. Also, find the matrix represented by $A^5-4A^4-7A^3+11A^2-A-10I$.

Technique

Cayley–Hamilton reduces all higher powers $A^k$ ($k\ge 2$) to linear combinations of $A$ and $I$.

Solution

Step 1 — Characteristic polynomial and Cayley–Hamilton

Cayley–Hamilton: $A^{2}-4A-5I=0$.

Verify directly:

Step 2 — Inverse via Cayley–Hamilton

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

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

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

Step 4 — Evaluate $A^{5}-4A^{4}-7A^{3}+11A^{2}-A-10I$

Substitute each power as $\alpha A+\beta I$:

Sum:

• $A$-coefficient: $521-416-147+44-1=1$.
• $I$-coefficient: $520-420-140+55-10=5$.

Result: $A+5I$.

Answer