← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q4a — Step-by-Step Solution

15 marks · Section A

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

Question

State the Cayley-Hamilton theorem. Use this theorem to find $A^{100}$, where $A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$.

Technique

State CH; get $p(\lambda)=(\lambda-1)^2(\lambda+1)$; write $\lambda^{100}\equiv a\lambda^2+b\lambda+c\pmod{p}$ and pin $a,b,c$ using the (repeated) eigenvalues, including the derivative condition at the double root.

Solution

Step 1 — Statement of the Cayley–Hamilton theorem

Cayley–Hamilton theorem. Every square matrix satisfies its own characteristic equation. If $A$ is $n\times n$ and $p(\lambda)=\det(\lambda I-A)$ is its characteristic polynomial, then $p(A)=O$ (the zero matrix).

Step 2 — Characteristic polynomial of $A$

Expanded: $p(\lambda)=\lambda^3-\lambda^2-\lambda+1$. By Cayley–Hamilton,

Step 3 — Reduce $A^{100}$ via the characteristic polynomial

Divide $\lambda^{100}$ by $p(\lambda)=(\lambda-1)^2(\lambda+1)$:

Since $p(A)=O$, we will get $A^{100}=aA^2+bA+cI$. Determine $a,b,c$ from the roots of $p$:

• $\lambda=1$ (double root): substitute $\lambda=1$:

• Differentiate and put $\lambda=1$ (double root forces the derivative to match): $100\lambda^{99}=q'(\cdot)\,p+q\,p'+(2a\lambda+b)$, and $p(1)=p'(1)=0$, so

• $\lambda=-1$:

From (i) and (iii): $(a+b+c)-(a-b+c)=2b=0\Rightarrow b=0$. Then (ii) gives $2a=100\Rightarrow a=50$, and (i) gives $c=1-50=-49$.

Step 4 — Compute $A^2$ and assemble

Then

Answer