How do you factor #((1, 2, -3), (0, 1, 3), (0, 0, 1))# into a product of elementary matrices?

Question

2581 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-06T20:57:08

#((1, 2, -3),(0, 1, 3),(0, 0, 1)) = ((1, 2, 0),(0, 1, 0),(0, 0, 1))((1, 0, -9),(0, 1, 0),(0, 0, 1))((1, 0, 0),(0, 1, 3),(0, 0, 1))#

Given:

#((1, 2, -3),(0, 1, 3),(0, 0, 1))#

We can describe the process of making this matrix into the identity matrix as follows:

(1) Subtract #2 xx "row2"# from #"row1"# to get:

#((1, 0, -9),(0, 1, 3),(0, 0, 1))#

(2) Add #9 xx "row3"# to #"row1"# to get:

#((1, 0, 0),(0, 1, 3),(0, 0, 1))#

(3) Subtract #3 xx "row3"# from #"row2"# to get:

#((1, 0, 0),(0, 1, 0),(0, 0, 1))#

Reversing the steps and expressing the row operations as elementary matrices we arrive at the following product:

#((1, 2, 0),(0, 1, 0),(0, 0, 1))((1, 0, -9),(0, 1, 0),(0, 0, 1))((1, 0, 0),(0, 1, 3),(0, 0, 1))#