JEE Advanced 2022 Paper 1 · Mathematics · Matrices and Determinants · hard

Let A=beginpmatrix2&1&0\\0&3&1\\0&0&1endpmatrix and suppose A^5=pA^2+qA+rI for real numbers p,q,r. Find p-q+r.

Numerical answer

Let

A=\begin{pmatrix}2&1&0\\0&3&1\\0&0&1\end{pmatrix}

and suppose

A^5=pA^2+qA+rI

for real numbers

p,q,r

. Find

p-q+r

Show answer and solution

Answer: 479

The characteristic polynomial of the triangular matrix

A

(\lambda-1)(\lambda-2)(\lambda-3)=\lambda^3-6\lambda^2+11\lambda-6

. By Cayley-Hamilton the reduction of

\lambda^5

modulo it is

p\lambda^2+q\lambda+r

with equality at

\lambda=1,2,3

(distinct roots):

p+q+r=1

4p+2q+r=32

9p+3q+r=243

. Solving:

p=90,\ q=-239,\ r=150

(verified:

A^5-90A^2+239A-150I=0

). Then

p-q+r=90+239+150=479

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake