Question 87122
<pre><font size = 4 color = "darkred"><b>
Prove or disprove whether the following matrices are 
equivalent or not. Explain how you came up with your answer. 

[1 2 3]
[0 1 1] 

[3 2 5]
[5 4 9]

We must get both of these into "row reduced echelon form"

Let's get the first one to row reduced echelon form:

[1 2 3]
[0 1 1]

Get a 0 where the 2 is by multiplying the bottom row by -2
and adding it to 1 times the top row, then restoring 
the bottom row:

 1[1 2 3]
-2[0 1 1]

[1 0 1]
[0 1 1]

That is in row reduced echelon form because:
1. the leading 1 in the bottom row is farther
   to the right than the leading 1 in the top
   row.
2. Each of the leading 1's are the only non-zero
   numbers in the columns they are in. 

Now, let's get the second one to row reduced echelon form:

[3 2 5]
[5 4 9]

Get a 0 where the 5 on the bottom row is by multiplying the 
top row by -5 and adding it to 3 times the bottom row, then 
restoring the top row:

-5[3 2 5]
 3[5 4 9]

[3 2 5]
[0 2 2]

Get a 0 where the 2 on the top row is by multiplying the 
bottom row by -1 and adding it to 1 times the top row, then 
restoring the bottom row:

 1[3 2 5]
-1[0 2 2]

[3 0 3]
[0 2 2]

Get a 1 where the first 3 in the top row is by dividing the
top row through by 3.

Get a 1 where the first 2 in the bottom row is by dividing the
bottom row through by 2.

[3 0 3]÷3
[0 2 2]÷2

[1 0 1]
[0 1 1]

So the given matrices are equivalent because the have the
same row reduced echelon form, namely

[1 0 1]
[0 1 1] 

Note: <i>It didn't apply here but in some matrices, to have 
row reduced echelon form, all all-zero rows must appear
at the bottom of the matrix.  This can be accomplished
by swapping rows.</i>

Edwin</pre>