Thread: Row-equivalent matrices

Hi,
I am having trouble with this question:
With these two matrices,

$\displaystyle \left(\begin{array}{ccc}2&0&1\\1&3&1\\5&-3&2\end{array}\right)$ and $\displaystyle \left(\begin{array}{ccc}1&a&b\\a&1&c\\b&c&a\end{ar ray}\right)$

What numbers a, b, and c make them row-equivalent

I know that if their row equivalent, they have to have the same reduced row echelon form. So I tried approaching it from this angle, but I can't really transform the second matrix to a reduced-row echelon form

Any suggestions?

normally, that's how we do it. But alike yourself, i'm having trouble figuring out RREF humm

maybe we can supposed to use some other facts about row-equivalent matrices?

if someone can help out, which i'm sure of, i would be interested in technique of approaching this problem as well. Thanks.

here is an approach that might do the job.

$\displaystyle A= \left(\begin{array}{ccc}2&0&1\\1&3&1\\5&-3&2\end{array}\right)$ and $\displaystyle B= \left(\begin{array}{ccc}1&a&b\\a&1&c\\b&c&a\end{ar ray}\right)$

use linear combination facts (swamp, pivot, and scale) to match entries of matrices A to that of B.

make...

{ note: r = row; c = column }

A (r1c1) = 1 b/c B (r1c1)

A (r2c2) = 1 b/c B (r2c2)

A (r2c1) = A (r2c1) = A (r3c3) b/c B (r2c1) = B (r2c1) = B (r3c3)

A (r3c1) = A (r1c3) b/c B (r3c1) = B (r1c3)

and

A (r2c3) = A (r3c2) b/c B (r2c3) = B (r3c2)



Unless i'm really gone mental (which could be the case this night lol) the above technique should be valid.

After all, as you said, two matrices A and B are row-equivalent iff RREF of matrices A and B are the same.

We can take matrix A, find it's RREF, which is also the RREF of matrix B that is row-equivalent to A, and then go back to B humm


P.S. i'll give those #s a try later tomorrow.

That seems valid, but in terms of finding the right elementary operations, it seems like it would be guesswork. I'm not sure exactly where to start since there are so many possible operations.

By the way, the RREF of matrix A is

$\displaystyle \left(\begin{array}{ccc}1&0&1/2\\0&1&1/6\\0&0&0\end{array}\right)$

unfortunately you're right...too many possibilities to consider.