The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+1 vote
39 views

asked in Linear Algebra by Active (3.3k points)
retagged by | 39 views
+1

Eigen value of A:  1 , -1 and x

18 is eigen value of $A^2 +3A$==>A's 3rd Eigen value of A be either 3 or -6.

|A|=product of eigen value=1*-1*(3||6) !=0 So invertible and for |$A^2 +3A $ |=18*4*(-2) which is also non zero.

Hence Both invertible.

can  u @srestha @Shaik Masthan plz verify.

0

@Abhisek Tiwari 4

yes correct I think :)

0
Thanks.. ;)

Please log in or register to answer this question.

Related questions

+6 votes
1 answer
5
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,814 questions
54,522 answers
188,364 comments
75,384 users