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Chapter 7: Q51E (page 346)

Find the characteristic polynomial of the matrix A=[00a10b01c]

where a, b, and c are arbitrary constants.

Short Answer

Expert verified

Characteristic polynomial of the matrix isfA(位)=-位3+肠位2+产位+a

Step by step solution

01

Characteristic equation 

Consider an n 脳n matrix A and a scalar 位. Then 位 is an eigenvalue5 of A.

det(A-位濒)=0

This is called the characteristic equation (or the secular equation) of matrix A.

02

Solution of the problem

We solve:

A=00a10b01c

Characteristic polynomial is given bydet(A-位濒)=0

(A-位濒)=0-位0a10-位b01c-位=-位0a1-位b01c-位det(A-位濒)=-位(-位(c-位)-b)-0+a(1-0)

Here,

0=位2(c-位)+产位+a

Therefore,

-位3+肠位2+产位+a=0fA(位)=-位3+肠位2+产位+a

The final result isfA(位)=-位3+肠位2+产位+a

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