A = [[1,2,11,12,-33],[-3,-5,-28,-32,87],[-1,-3,-16,-18,49]] Given a row reduction of A B = [[1,0,1,0,-1],[0,1,5,0,-4],[0,0,0,1,-2]] find an invertible matrix U such that B = UA.

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The matrix  `U` is the product of some particular set of elementary matrices. An elementary matrix can be reached from the identity matrix of the same dimension by an elementary row or column operation.

Each operation in a row reducing process can be represented by an elementary matrix.

Working through the row reduction from A to B we have

i) row 2 `->` row 2 + 3 row 1

ii) row 3 `->` row 3 + row 1

After these two operations we are left with

`((1,2,11,12,-33),(0,1,5,4,-12),(0,-1,-5,-6,16))`

We then proceed with

iii) row 3 `->` row 3 + row 2

iv) row 3 `->` (-1/2) x row 3

v) row 2 `->` row 2 - (4 x row 3)

vi) row 1 `->` row 1 - (12 x row 3)

We are then left with

`((1,2,11,0,-9),(0,1,5,0,-4),(0,0,0,1,-2))`

Finally we reduce by

vii) row 1 `->` row 1 - (2 x row 2)

Putting all this together

`U = ((1,-2,0),(0,1,0),(0,0,1))((1,0,-12),(0,1,0),(0,0,1))((1,0,0),(0,1,-4),(0,0,1))((1,0,0),(0,1,0),(0,0,-1/2)) `

` times ((1,0,0),(0,1,0),(0,1,1))((1,0,0),(0,1,0),(1,0,1))((1,0,0),(3,1,0),(0,0,1))`

Multiplying pairs together we get

` ` `U = ((1,-2,-12),(0,1,0),(0,0,1))((1,0,0),(0,1,2),(0,0,-1/2))((1,0,0),(0,1,0),(1,1,1))((1,0,0),(3,1,0),(0,0,1))`

`= ((1,-2,2),(0,1,2),(0,0,-1/2))((1,0,0),(3,1,0),(4,1,1))``= ((3,0,2),(11,3,2),(-2,-1/2,-1/2))`

Check

`UA =((3,0,2),(11,3,2),(-2,-1/2,-1/2))((1,2,11,12,-33),(-3,-5,-28,-32,87),(-1,-3,-16,-18,49))`

`= ((1,0,1,0-1),(0,1,5,-4),(0,0,0,1-2))`

Therefore

`U = ((3,0,2),(11,3,2),(-2,-1/2,-1/2))`  answer (each elementary matrix component of U is invertible, so U is invertible)

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