Can anyone please help me solve this question part a ? I solved it but the equation i m getting is i think wrong because when i put in factor no and its gives me whole equation equal to zero that s how i will get my eigen values.

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λ^3-15 λ^2+65 λ-120=0

please have a look if this equation is right?

1 Answer
Mar 1, 2018

#lambda=4 ,5 , 6#

Explanation:

#bbA=[(1,2,3),(-2,6,2),(-4,2,8)]#

We need to solve the system:

#bb(AX)=lambdabbX#

#bb(AX)-lambdabbX=0#

Factoring:

#(bb(A)-lambdabbI)bbX=0#

Where #bbI# is the identity matrix.

This is a homogeneous system, so in order to have a non zero solution the determinant must equal zero.

#|bbA-lambdabbI|=0#

#[(1,2,3),(-2,6,2),(-4,2,8)]-lambda[(1,0,0),(0,1,0),(0,0,1)]=|(1-lambda,2,3),(-2,6-lambda,2),(-4,2,8-lambda)|#

I will have to do this in parts due to space:

#[(1-lambda)[(6-lambda)(8-lambda)]-(4)]=44-58lambda+15lambda^2-lambda^3#

#(-2)[(-2)(8-lambda)-(-8)]=16-4lambda#

#(3)[(-4)-((-4)(6-lambda))]=60-12lambda#

#44-58lambda+15lambda^2-lambda^3+16-4lambda+60-12lambda#

#=-lambda^3+15lambda^2-74lambda+120#

#=lambda^3-15lambda^2+74lambda-120=0#

#(lambda-5)(lambda-6)(lambda-4)=0#

#lambda=4 ,5 , 6#