MapAt a function at specific positions

Question

I'm trying to solve for $p3$ in 3p3-5 = p1 + p2 at positions where p1*p2 == 0:

deltap = 10^(-2)

tabletry = Table[p1*p2, {p1, 0, 2}, {p2, 0, 2}]  // MatrixForm

tabletest = Table[{If[p1*p2 == 0, True, False  ]}, {p1, 0, 2}, {p2, 0, 2}] // MatrixForm

$$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \\ \end{array} \right)$$

$$\left( \begin{array}{ccc} {(True)}\ & (True) & (True) \\ \ (True) & (False) & (False) \\ \ (True) & (False) & (False) \\ \end{array} \right) $$

(Array[# #2 /. {(0) -> Style[Solve[3 p3 - 5 == # + #2, p3][[1, 1, -1]], Red], _ :> 
   0} &, {3, 3}, 0]) // MatrixForm

EDIT

Now I'm trying to up the level of code, trying to evaluate the difference in p3 following at true entries. Full code:

deltap = 10^-2;

tabletry = (Table[p1*p2, {p1, 0, 2}, {p2, 0, 2}] ) // MatrixForm

tst = ( Map[{# ==  0 } &, tabletry, {-1}] ) // MatrixForm    

eqn1 = 1/p3 - 5p3 = p1 + p2 

eqn2 = 1/p3 - 5p3 + deltap = p1 + p2

diff = p3 /. FindRoot[eqn2, {p3,1}] - p3 /. FindRoot[eqn1,{p3,1}]

(Array[# #2 /. {(0) -> Style[Evaluate[diff[#, #2]][[1, 1, -1]], Red], _ :> 
       0} &, {3, 3}, 0]) // MatrixForm

Answer 1

For your second question, you could use

(tabletry = Table[p1*p2, {p1, 0, 2}, {p2, 0, 2}]) // MatrixForm;
(* or  (tabletry = Array[# #2 &, {3, 3}, 0]) // MatrixForm; *)
(tst = Map[{# == 0} &, tabletry, {-1}]) // MatrixForm 

or

(tst2 = Array[# #2 /. {0 -> {True}, _ -> {False}} &, {3, 3}, 0]) // MatrixForm

Note the parantheses wrapping the definitions -- without those you would be setting the lhs to MatrixForm[rhs]

For the first part, assuming you want to leave the non-zero entries untouched, you could use

(mybestwildguessaboutexpectedoutput = 
   Array[# #2 /. (0) -> Style[Solve[3 p3 - 5 == # + #2, p3][[1, 1, -1]], 
                              Red, Bold] &, {3, 3}, 0]) // MatrixForm

You could also use

MapIndexed[# /. (0) -> Solve[3 p3 - 5 == Plus @@ (#2 - 1), p3][[1, 1, -1]] &, tabletry, {2}]

Answer 2

Your p3 definitions seem different. The following uses 3p3 -5=p1+p2.

set = Tuples[Range[0, 2], 2];
set /. {x_, y_} :> (x + y + 5)/3. /; x y == 0

yields:

{1.66667, 2., 2.33333, 2., {1, 1}, {1, 2}, 2.33333, {2, 1}, {2, 2}}

or if you wish to couple results and {p1,p2}:

set /. {{x_, y_} :> 
   Rule[{x, y}, (x + y + 5)/3.] /; x y == 0, {x_, y_} :> 
   Sequence[] /; x y != 0}

yields:

{{0, 0} -> 1.66667, {0, 1} -> 2., {0, 2} -> 2.33333, {1, 0} -> 
  2., {2, 0} -> 2.33333}

Answer 3

I post this as another answer (using again $ 3p3=p1+p2-5$) if the matrices are the main aim:

sa0 = SparseArray[{i_, j_} :> (i - 1) (j - 1), {3, 3}] // MatrixForm;
sa = SparseArray[{{i_, 
        j_} /; ((i - 1) (j - 1) != 0) :> (i - 1) (j - 1), {i_, 
        j_} /; ((i - 1) ( j - 1) == 0) :> (i + j + 5)/3.}, {3, 3}] // 
   MatrixForm;
Row[{sa0, "\[RightArrow]", saf}] 

UPDATE

If you are just dealing with linear combinations, and your equations can be represented as $r p3 = m p1+ n p2+s$ then (borrowing from kguler and comments re desired output):

h[0, {a_, b_}, func_, v_] := 
 Style[v /. First@Solve[func, v], Bold, Red]
h[x_?(# != 0 &), __] := "x"

then you can use MapIndexed:

Here demonstrated using random quadruplet of integers:

Column[Function[{u, v, w, t}, 
   u p3 == {p1, p2}.{v, w} + 
      t -> (MapIndexed[h[#1, #2, u p3 == #2.{v, w} + t, p3] &, 
       SparseArray[{i_, j_} :> (i - 1) (j - 1), {3, 3}], {2}] // 
      MatrixForm)] @@@ RandomInteger[{1, 10}, {10, 4}], Frame -> All]