I have a very large n x n matrix from which I want to take a row i from and substitute every element with 0. When I try the below command however, the ith row is not replaced by a list of n zeros but a single zero, which breaks the matrix.

ReplacePart[A, {i -> {0}}]

Is there a neat way of substituting that {0} with some command that produces an automatic list of (length[A]) number of 0s?

  • $\begingroup$ For in place modification (no copy), maybe A[[4,All]]*=0 $\endgroup$
    – user1066
    Jan 15 at 15:53
  • 2
    $\begingroup$ Another possibility is to use Dot, e.g. DiagonalMatrix[1 - UnitVector[Length[A], i]].A $\endgroup$
    – Carl Woll
    Jan 15 at 16:03
A = Array[a, {10, 10}];

i = 4;
b = ReplacePart[A, {i, _} -> 0];

b // MatrixForm

enter image description here

You can also do

c = A; c[[i, ;;]] = 0;

d = MapAt[0 & /@ # &, A, {i}];

e = MapAt[0 &, A, {i, All}];

b == c == d == e

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