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  •  » Matrix Algebra - Reflection of Vector and Leading one's of rref matrix

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Topic review (newest first)

MathsIsFun
2005-10-21 07:18:59

"modueration exam" ? smile

Anyway, tell us what you need, we may be able to point you in the right direction.

priya
2005-10-21 04:30:13

i have got modueration exam coming up and i can't find the work that i need. i'm doing foundation work at the moment i really need your help.

Thanks,
bye

kybasche
2005-10-06 07:56:12

For the first problem, I think the one I was missing was

[0,1:0,0:0,0]

Anyone have thoughts on the second problem?

~Derek

kybasche
2005-09-23 14:02:26

Hey all, lovely forum!  I've got 2 problems, both seem to be pretty straight forward, but I'm getting stuck.  I've used the notation for matrices that my calculator accepts... a comma seperates values in the same column, and a colon denotes the start of a new row, I also seperated rows with spaces to make it a bit easier to read

1)  (this is paraphrased from Linear Algebra w/ Applications by Otto Bretscher)

two nxm matrices in reduced row-echelon form are the the same type if they have the same number of leading 1's in the same locations... i.e. [*1,2,0 : 0,0,*1] and [*1,3,0 : 0,0,*1] --- stars indicate corresponding leading 1's

How many types of 3x2 matrices in rref are there?

I get 3... [1,a : 0,0 : 0,0]    [1,0 : 0,1 : 0,0]  and the [0,0 : 0,0 : 0,0]
However, the book's answer is that there are 4 types... am I missing something??

2)   (again paraphrased from the same book)

consider matrix A = [a,b : b,-a]  where a^2+b^2=1

Find two nonzero perpendicular vectors 'v' and 'w' such that A*v=v and A*w=-w
Solve for the vectors in terms of 'a' and 'b'

For this one, the matrix 'A' is a reflection transformation, and in order for the reflection of 'v' to be equal to 'v,' I would imagine that 'v' has to be parallel to the line about which the reflection is taking place.  I can't seem to find a way to get to the answer though.

The answer is given in the book as

v = [b : (1-a)]
w = [-b : (1+a)]



Sorry that this was so long smile  Any help on either of these would be very much appreciated.  Thanks a bunch!

~Derek

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