thanks hallsofevy that was helpful!
since x and y could be anything
so the possible eigen vectors corresponding to eigen value 0 are
[1 0 0],[0 1 0] or [1 1 0]
please tell me am i correct now?
hey anyone please can tell me this one:
A=lim(e^(1/n*logn)) (n tends to 0)
i took log on both sides and then by using l hospital rule
i arrive at lim(-1/n) (n tends to 0)
can't solve further...please help
sorry hallsofevy, the correct answer is [a 0 0]
but after solving the eqn [A-lamdaI][X]=0 or AX=lamdaX
i got az+0+0 = 0
so how to proceed further and arrive at the correct answer?
hey sourabh, i did try to solve the question.
i got eigen values as 0, 0 , 0
and after using [A-lambdaI]X=0
i am getting 0X1+0X2+aX3=0
which makes eigen vector as [0 0 0 ]
whereas the answer is [0 0 a]