# Suppose that: E_6 [[1,3,4],[2,2,3],[-5,4,4]] = [[1,3,4],[2,2,3],[-2,13,16]] Find E_6.

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`E_6` is an elementary matrix. That is it is a matrix that can be reached from the identity matrix of the same dimension by an elementary operation.

We have that

`E_6 ((1,3,4),(2,2,3),(-5,4,4)) = ((1,3,4),(2,2,3),(-2,13,16))`

We can see that the first two rows are preserved and the third row is neither a multiple of itself nor a multiple of the other two rows. Eliminating these possibilities we are left with the fact that row 3 must be a linear combination of the rows.

Write `E_6 = ((1,0,0),(0,1,0),(a,b,c))`. Then

`((1,0,0),(0,1,0),(a,b,c))((1,3,4),(2,2,3),(-5,4,4)) = ((1,3,4),(2,2,3),(-2,13,16))` so that

` ` `((1,2,-5),(3,2,4),(4,3,4))((a),(b),(c)) = ((-2),(13),(16))`

Reduce by row reduction:

`((1,2,-5),(0,-4,19),(0,-5,24))|((-2),(19),(24))` , ```((1,2,-5),(0,1,-19/4),(0,-5,24))|((-2),(-19/4),(24))`

`((1,2,-5),(0,1,-19/4),(0,0,24-95/4))|((-2),(-19/4),(24-95/4))` ,

`((1,2,-5),(0,1,-19/4),(0,0,1))|((-2),(-19/4),(1))` , `((1,2,0),(0,1,0),(0,0,1))|((3),(0),(1))` , `((1,0,0),(0,1,0),(0,0,1))|((3),(0),(1))`

Therefore `a=3` and `c=1` so that `E_6` is reached from the identity matrix by replacing row 3 by (3 x row 1) + row 3. We then have that

`E_6 = ((1,0,0),(0,1,0),(3,0,1))`

**Answer. Note this could have been obtained by inspection but may have been difficult to spot as there were many possible combinations.**