J: Gauss-Jordan elimination - j

Question

Task to code Gauss-Jordan method of solving linear system of algebraic equations is an exercise that I've selected to advance in learning J. System is Ax=b, where A is n-by-n matrix, b and unknown x are n-vectors. Firstly, I've started with the simplest form with control structures:
gj0 =: dyad :0 NB. usage is same to %.
y=.y,.b
for_d.i.#y do.
for_r.i.#y do.
if.r=d do.continue.end. NB. do not eliminate d'th row
t=.%/ (<"1(r,d),:(d,d)) { y
for_c.d}.>:i.#y do.
y=.(((<r,c){y)-(t*(<d,c){y)) (<r,c)} y
end.
y=.0 (<r,d)} y NB. ensure zero
end.
end. NB. now A is diagonal but not identity matrix, so:
x=.{:"1 y NB. x = b
for_r.i.#y do.
x=.((r{x)%(<r,r){y) r} x NB. divide by coefficients on diagonal
end.
)
Ab =: (".;._2) 0 :0
0.25 _0.16 _0.38 0.17
0.19 _0.22 _0.02 0.41
0.13 0.08 _0.08 _0.13
0.13 _0.1 _0.32 0.65
)
b =: 0.37 0.01 0.01 1.51
(,.".&.>)('A';'b';'gj0 A,.b';'b %. A')
┌────────┬──────────────────────┐
│A │0.25 _0.16 _0.38 0.17│
│ │0.19 _0.22 _0.02 0.41│
│ │0.13 0.08 _0.08 _0.13│
│ │0.13 _0.1 _0.32 0.65│
├────────┼──────────────────────┤
│b │0.37 0.01 0.01 1.51 │
├────────┼──────────────────────┤
│b gj0 A │_1 3 _2 2 │
├────────┼──────────────────────┤
│b %. A │_1 3 _2 2 │
└────────┴──────────────────────┘
Correct! Next I've decided to get rid of as many control structures as possible:
gj1 =:dyad :0
y=.y,.b
for_d.i.#y do.
for_r.d ({.,]}.~[:>:[) i.#y do. NB. for indices without d
t=.%/ (<"1(r,d),:(d,d)) { y
y=.((r{y)-(t*d{y)) r}y NB. no need to iterate for each column
y=.0 (<r,d)} y
end.
end.
({:"1 y)%(+/}:"1 y) NB. b divide by sum of each item of A (drop zeroes)
)
b gj1 A
_1 3 _2 2
OK, Now I can try to translate for_r.-loop into tacit form... but it seems like it will look more cumbersome and I think that I'm at a wrong way -- but what's study without mistakes? I really want to code Gauss-Jordan method tacitly to:
exercise in J coding
see if it is better in performance
try to understand code a few weeks later :)
Help me please write it to the end or point out a better approach.

Answer 1

Thanks to Eelvex, who advised me to look in addons/math/misc/linear.ijs, I've concluded the task with this nice code:
gj=: monad :0
I=. i.#y
for_i. I do. y=. y - (col - i=I) */ (i{y) % i{col=. i{"1 y end.
)
gj Ab
1 0 0 0 _1
0 1 0 0 3
0 0 1 0 _2
0 0 0 1 2
It has taken some time to understand verb pivot in linear.ijs - but pencil-paper method helps.