How to remove extra parentheses in a matrix [closed]

I have the following matrix with Dimensions[A] = {5}. What is the easiest way to convert it to $$8 \times 8$$ matrix?

A = {{{1, 1, 1, 1, 1, 1, 1, 1}}, {{1, 3, 2, 8, 6, 5, 9, 2}},
{{1, 8, 9, 8, 2, 7, 7, 5}, {7, 9, 6, 1, 8, 7, 6, 3}},
{{2, 9, 5, 5, 1, 7, 9, 7}, {9, 4, 8, 3, 5, 6, 2, 6},
{1, 2, 5, 3, 3, 8, 5, 2}, {4, 4, 2, 3, 6, 8, 7, 7}},
{{6, 5, 9, 3, 3, 5, 8, 2}, {4, 1, 4, 7, 4, 4, 8, 2},
{4, 8, 8, 6, 6, 4, 6, 3}, {8, 4, 1, 6, 6, 3, 2, 1},
{4, 9, 6, 3, 6, 9, 5, 6}, {9, 2, 5, 8, 6, 3, 5, 6},
{4, 2, 2, 2, 2, 5, 8, 6}, {6, 2, 2, 6, 9, 9, 5, 2},
{6, 4, 7, 6, 8, 9, 1, 8}}};

• Let me first make sure I understand if this is what you want. ArrayReshape[Flatten[A], {8, 8}]; Dimensions[%] gives {8,8} but your original matrix has 136 entries, not 64? Commented Jun 8, 2023 at 14:06
• @Nasser Yes that's what I was looking for, thanks. I wish you left the answer I would have accepted it. Commented Jun 8, 2023 at 14:06
• OK, it put the answer back. I just was not sure if this is what you wanted as the number of entries is different, that is why. Commented Jun 8, 2023 at 14:07
• Dimensions /@ A evaluates to {{1, 8}, {1, 8}, {2, 8}, {4, 8}, {9, 8}} and Length@Flatten@A evaluates to 136 Which 64 elements are supposed to be in the final 8x8 matrix? Commented Jun 8, 2023 at 14:11
• Very strong recommendation: Post the MINIMAL data/code that illustrates your problem. Do we really need to see every number to 12 digits?? The simpler version will help you see the answer yourself, oftentimes. Commented Jun 8, 2023 at 14:44

If I understand you right

ArrayReshape[Flatten[A], {8, 8}];
Dimensions[%]


btw, the title of you question says

How to remove extra parentheses in a matrix

If this is what you really wanted, then the command is

Flatten[A, 1]
Dimensions[%]


So may be you wanted the first 8 rows of the above only and discard the rest of the rows? In this case, this also works

Flatten[A,1];
%[[1;;8,All]];
Dimensions[%]


Your list seems to have the structure:

{Union@Flatten[A], Length[B=Flatten[A,1]], Length/@#&/@A}
(* {{1, 2, 3, 4, 5, 6, 7, 8, 9}, 17,
{{8}, {8}, {8, 8}, {8, 8, 8, 8}, {8, 8, 8, 8, 8, 8, 8, 8, 8}}} *)


At the lowest level you have $$17$$ Lists of length $$8$$ which are of integers from $$1$$ to $$9$$. These are collected in matrix $$B$$. Some $$8$$ rows of this matrix can be collected to form an $$8 \times 8$$ matrix. The most direct way is to do this is Take[B,8] for the first $$8$$ rows but any other $$8$$ rows from B can be used. For example Take[B,-8] gives the last $$8$$ rows. It depends on what you want to extract from A.

If you want the first eight 8-vectors (as in @Nasser's answer):

B = Take[
Level[A, {-2}], (* all row vectors *)
8]
B // Dimensions
(*
{{...}}
{8, 8}
*)