Consider the following row reduction.

A = [[0 0 1],[0 3 0],[1 0 2]] --> [[1 0 2],[0 3 0],[0 0 1]] --> [[1 0 2],[0 1 0],[0 0 1]] --> [[1 0 0],[0 1 0],[0 0 1]]

Using this row reduction, write A and A^-1 as products of elementary matrices.

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An elementary matrix is one that can be reached from the identity matrix I by a single elementary row operation, like row addition or row switching.

Therefore

`A = ((0,0,1),(0,3,0),(1,0,2)) = ((0,0,1),(0,1,0),(1,0,0))((1,0,2),(0,3,0),(0,0,1)) `

`= ((0,0,1),(0,1,0),(1,0,0))((1,0,0),(0,3,0),(0,0,1))((1,0,2),(0,1,0),(0,0,1))`

so `A` is a product of the elementary matrices `R_1 harr R_3`, `R_2-> 3R_2` and `R_1 -> R_1 + 2R_3` where `R_i` denotes row `i`

` ``A^(-1)` will simply be the inverse of this, ie

`R_1 -> R_1 -2R_3`, `R_2 -> 1/3R_2` and then `R_1 harr R_3` so that

`A^(-1) = ((1,0,-2),(0,1,0),(0,0,1))((1,0,0),(0,1/3,0),(0,0,1))((0,0,1),(0,1,0),(1,0,0))`

Check. This gives

`A^(-1) = ((-2,0,1),(0,1/3,0),(1,0,0))` so that

`A A^(-1) = ((0,0,1),(0,3,0),(1,0,2))((-2,0,1),(0,1/3,0),(1,0,0)) = ((1,0,0),(0,1,0),(0,0,1)) = I_3`

**A can be written as **`R_1 harr R_3` , `R_2 -> 3R_2` **and **`R_1 -> R_1 + 2R_3`.

**A^(-1) can be written as the inverse of the product of elementary matrices that gives A:**

**`R_1 -> R_1 - 2R_3`, `R_2 -> 1/3R_2` and then `R_1 harr R_3` **

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