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#1 2007-07-21 09:50:45

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Linear Algebra: Row echelon form

given the matrix A =

| 1  1   0  1   4 |
| 2  0   0  4   7 |
| 1  1  1  0    5 |
| 1 -3 -1 -10 a |
determine the value of a such that, in row echelon form of A, the stairstep reaches the bottom.[/b]

I guess they mean the bottom row cannot be all zero. I had no clue how to do this so i tried reducing it to row echelon form, treating a as an unnamed contstant. No matter what i did, i never ended up with all a's in the bottom row so it seems like it wouldn't matter what a is. But the answer says a = 1.

Any ideas?

Last edited by mikau (2007-07-21 10:04:43)


A logarithm is just a misspelled algorithm.

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#2 2007-07-21 12:24:28

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Linear Algebra: Row echelon form

This is just fiddling, but I put in 1 for a and tried to solve for the 4 variables.
And I got these values:   3.5  0.5   1   0  for the variables for each column.
Don't know what this means, but hope you like it.


igloo myrtilles fourmis

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#3 2007-07-21 17:01:36

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Linear Algebra: Row echelon form

thats interesting, it may have something to do with the answer. But I don't know what.


A logarithm is just a misspelled algorithm.

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#4 2007-07-21 18:18:24

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Linear Algebra: Row echelon form

Reducing the entire matrix I get the bottom row as:

0 0 0 -16 a-1

But like you, I'm not entirely sure what "stairstep reaches the bottom" means.  Though setting a to 1 would make that last column 0.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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