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Mathematics 18 Online
OpenStudy (anonymous):

i need to find the 100th power of the following 3x3 matrix...... 2,0,0 0,0,-1 0,3,0 my question is can interchange row 2 and 3 before calculation. because that will make it diagonal and its easy to calculate the power of a diagonal matrix....

OpenStudy (asnaseer):

yes - work out which matrix you need to multiply your matrix by in order to swap rows 2 and 3.

OpenStudy (asnaseer):

then note what happens to that matrix when you square it

OpenStudy (asnaseer):

i.e. lets call your matrix A, then find a matrix B such that: C = BA and C contains A with rows 2 and 3 swapped

OpenStudy (asnaseer):

then note what \(B^2\) is equal to

OpenStudy (asnaseer):

and take advantage of that

OpenStudy (anonymous):

do determinants

OpenStudy (anonymous):

you mean in this case multiply with 1 0 0, 0 0 1, 0 1 0 right??

OpenStudy (asnaseer):

yes

OpenStudy (anonymous):

so its alright to do that??? ii mean i was thinking it might change the property of original matrix A and so can not be performed

OpenStudy (asnaseer):

hang on...

OpenStudy (asnaseer):

I made a mistake up there, what you actually need to do is to find a matrix P such that: P^-1 A P = D where D is a diagonal matrix.

OpenStudy (asnaseer):

and then make use of: A^n = (P D P^-1)(P D P^-1)(P D P^-1)...

OpenStudy (anonymous):

ohhhh..... i know that....and i was trying to avoid that long path

OpenStudy (asnaseer):

which simplifies to: A^n = P D^n P^-1

OpenStudy (anonymous):

because it has a structure very close to diagonal

OpenStudy (anonymous):

so i was wondering whether a swqapping (if allowed) would save a huge time

OpenStudy (asnaseer):

no - because if you use my initial suggestion of: C = BA then you will get: A = B^-1 C so: A^n = (B^-1 C)(B^-1 C)... which doesn't really simplify

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