1 votes 1 votes $’X’$ 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? $X$ and $T$ both are invertible. $T$ is invertible but not $X$. $X$ is invertible but not $T$. None of them are invertible. GATE tbb-mockgate-3 engineering-mathematics linear-algebra matrix eigen-value + – Bikram asked Feb 9, 2017 retagged Sep 15, 2020 by ajaysoni1924 Bikram 400 views answer comment Share Follow See 1 comment See all 1 1 comment reply JashanArora commented Jan 16, 2020 reply Follow Share 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 votes 0 votes Please log in or register to add a comment.
Best answer 4 votes 4 votes 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 Rishabh Gupta 2 answered Feb 1, 2018 selected Dec 12, 2018 by Manoja Rajalakshmi A Rishabh Gupta 2 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes The matrix is invertible iff it has not 0 as eigen value. Bikram answered Feb 9, 2017 Bikram comment Share Follow See all 0 reply Please log in or register to add a comment.