Row-wise matrix multiplication

Question

I have a matrix $A=\{a_{ij}\}$ of dimensions $n$ rows by $m$ columns, and I would like to multiply each row of the matrix by a constant $b_i$, $i=1,\ldots,n$, and store the result in the same variable. In linear-algebra speak, I basically would like to do $A=\mathrm{diag}(b_1,\ldots,b_n) A$. However, I would like to avoid having to create a diagonal matrix off the vector $(b_1,\ldots,b_n)$, because $n$ is large and I wouldn't want to use any more RAM. Is there a different and less memory demanding way to do $A=\mathrm{diag}(b_1,\ldots,b_n) A$?

My approach is to go over the rows of $A$ and multiply each by the respective element of the vector $(b_1,\ldots,b_n)$. May be there is a more elegant way to do this? By "elegant" I mean faster yet not too memory-demanding.

m = 3;
n = 5;
A = Table[RandomInteger[{0, 5}], {i, m}, {j, n}];
b = Table[RandomInteger[{5, 10}], {i, m}];
A
Print[b];

(* Standard Linear-Algebraic Solution *)
b=DiagonalMatrix[b];
cc=Dot[b,A];
cc

(* Less memory demanding solution *)
For[i = 1, i <= m, i += 1,
  A[[i, All]] = b[[i]]*A[[i, All]];
  ];
A

Answer 1

This is also a kind of inner product, hence:

amat = Array[a, {3, 5}];
bvec = Array[b, 3];
ans = Transpose[Inner[Times, bvec, amat, List]]
MatrixForm[ans]

Or, numerically:

amat = RandomInteger[{-5, 5}, {3, 5}];
bvec = RandomInteger[{-5, 5}, 3];
ans = Transpose[Inner[Times, bvec, amat, List]];
MatrixForm[ans]

Answer 2

Oh, yes, there is a more elegant way!

Mathematica does element-by-element multiplication.

I think this does what you want:

A = Array[a, {3, 5}]

 (* {{a[1, 1], a[1, 2], a[1, 3], a[1, 4], a[1, 5]}, {a[2, 1], 
  a[2, 2], a[2, 3], a[2, 4], a[2, 5]}, {a[3, 1], a[3, 2], a[3, 3], 
  a[3, 4], a[3, 5]}} *)

 bvec = Array[b, {3}]

 (* {b[1], b[2], b[3]} *)

 bvec * A

 (* {{a[1, 1] b[1], a[1, 2] b[1], a[1, 3] b[1], a[1, 4] b[1], 
  a[1, 5] b[1]}, {a[2, 1] b[2], a[2, 2] b[2], a[2, 3] b[2], 
  a[2, 4] b[2], a[2, 5] b[2]}, {a[3, 1] b[3], a[3, 2] b[3], 
  a[3, 3] b[3], a[3, 4] b[3], a[3, 5] b[3]}} *)

I have used symbolic quantities in the test vector and matrix so you can see what is going on.

As a general rule, if you have a nested For loop in Mathematica, you are missing something about its capabilities.