in GATE retagged by
408 views
1 vote
1 vote

$’X’$ and $’T’$ are two square matrices. 

$’X’$ has eigen values $3, 0, 2$.   $’T’$ has eigen values $4$ and $1$.

Which of the following statements is CORRECT?

  1.     $X$ and $T$ both are invertible.
  2.     $T$ is invertible but not $X$.
  3.     $X$ is invertible but not $T$.
  4.     None of them are invertible.
in GATE retagged by
by
408 views

1 comment

A matrix has an eigenvalue 0 iff it is singular.

X has an eigenvalue 0. => X is singular. => X isn't invertible. (A matrix can either be singular or invertible. Not none, not both)


T has eigenvalues 4 and 1.

Hence, determinant of T = product of eigenvalues = 4.

Which is non zero.

=> Invertible.

0
0

2 Answers

4 votes
4 votes
Best answer

If any of the eigenvalues is 0, that means that the determinant of that matrix is zero. 

If the determinant is 0, then that matrix is singular, and it's inverse doesn't exist.

So, here (B)T is invertible but not X

selected by
1 vote
1 vote
The matrix is invertible iff it has not 0 as eigen value.
Answer:

Related questions

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

64.3k questions

77.9k answers

244k comments

80.0k users