how to calculate number of all square matrices where the entries are either 0 or 1?
491342
A square matrix is n by n, so it would have n^2 entries.  Each entry has 2 possible values: 0 or 1.  So, by the counting principle, the number of matrices with entries that are either 0 or 1 is 2^(n^2).
111022
7\n\norder them; if you have odd number of entries, take the middle one; if even - take average of two middle ones.\n\n\n\nand it's a statistics question
992764
64.\n\n1 cubed = 1\n2 cubed =8\n3 cubed =27\n4 cubed =64\n5 cubed = 125 etc.
1077320
My TI-83plus can do that...but it has to be an invertible matrix.  If the determinate is zero, then it can not be done
404639
you count the number of times each figure appears (this has been asked many other times):\nfirst row: 1\nsecond row: one 1: 11\nthird row: two 1: 21\nfourth row: one 2 one 1: 1211\nfifth row: one 1 one 2 two 1: 111221\nsixth row: three 1 two 2 one 1: 312211\n\n... and so on ...
194771
1. Write down a couple of Jordan blocks.\n\n2. Write down their squares, cubes, maybe even a fourth power.\n\n3. Gaze at each set of identity matrix, J, J^2, J^3, ... until you see a simple linear combination of the matrices that sums to the all zero matrix.  That is your minimal polynomial.\n\n4. Generalize.\n\n5. Now it is obvious. :-)\n\nAlternative:\n\n1.  Google for some combination of:\n\nminimal polynomial Jordan block\n"minimal polynomial" "Jordan block"\nJordan block minimal polynomial\n"Jordan block" "minimal polynomial"\n\n2. Read answer.\n\n3. Too bad, it still isn't obvious. :-(
252498
block matrices are those with equal number of rows and columns.\n\nex:\n\n1 1 1\n1 2 2 \n3 4 5\n\nthat's a 3x3 matrix.  same number of rows and columns.\n\nthey're the only matrices that can be multiplied. \n\nmatrices can never be divided by another matrix.
60125
you will always know how many squares to expect by using 2^n where n= # of alleles. In this case you have 2^8 or 2*2*2*2*2*2*2*2=256 squares.
1183592
If your matrix is not square, then you can compute a generalized inverse matrix.  There are some links below.  There are many generalized inverses for matrices (other than those that are square and invertible; those only have one generalized inverse, which is its inverse).  But there is only one Moore-Penrose inverse for any given matrix.\n\nedit: reetz, I have already reported you for copying.  I suggest you back off.
671808
4 sides and corners in a square
540085
(a-2)(a+2)  \n(difference of squares)
201316
This website may help you.  Good luck.\n\nhttp://www.sosmath.com/matrix/matrix.html
846422
Any matrix is a rectangular table of numbers and it used to describe linear equations. The horizontal lines in a matrix are called rows and the vertical lines are called columns. A matrix with m rows and n columns is called an m-by-n matrix (written m×n) and m and n are called its dimensions. The dimensions of a matrix are always given with the number of rows first, then the number of columns.\n\nMultiplication of two matrices is well-defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.  If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix (m rows, p columns) .\nnow how to make the multiplication process. assume any two matrices A 3x2 , B 2x2 first we check if we can get AB . and we can see the number of columns of A is the same as the number of rows of B. then the multiolication is possible . now you will multiply each row of A by each row of B to get the elements of AB ... If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix (m rows, p columns)\nso , AB is a 3x2 Matrix \n\nA=[ a b  , c d , e f ],     row1A= a b  ,   row2A= c d  , row3A= e f\nB=[g h , i j ] ---  row1B= g h  ,  row2B= i j\n\n\nAB= [ c1 c2   ,   c3 c4   ,   c5 c6  ]\n\nthe mult. process ,\n\nrow1A x row1B   will gives c1= (a x g) + ( b x h )\nrow1A x row2B   will gives c2= (a x i) + ( b x j)\n\nrow2A x row1B   will gives c3= (c x g) + ( d x h )\nrow2A x row2B   will gives c4= (c x i) + ( d x j )\n\nrow3A x row1B   will gives c5= (e x g) + ( f x h )\nrow3A x row2B   will gives c6= (e x i) + ( f x j )\n\nMatrix multiplication has the following properties:\n\n1- (AB)C = A(BC) for all k-by-m matrices A, m-by-n matrices B and n-by-p matrices C ("associativity"). \n\n2- (A + B)C = AC + BC for all m-by-n matrices A and B and n-by-k matrices C ("right distributivity"). \n\n3- C(A + B) = CA + CB for all m-by-n matrices A and B and k-by-m matrices C ("left distributivity"). \n\nIt is important to note that commutativity does not generally hold; that is, given matrices A and B and their product defined, then generally AB ≠ BA\n\n\nthats all.. i hope that i could help you :) you can contact me if you have any questions
23859
I run through the numbers from 1-9, looking at each to see where the constraints ensure that there is only one spot left to place an additional number.  To do this, realize that each row, each column and each block need exactly one of each number, so starting with 1, I look for combos with limited possibilities.\n\nIf the blocks 1,2 have a 1 in them, then 3 should be easy to find, look at blocks 6,9 to see if they have a 1, and mentally cross out the rows and columns that those 1's eliminate and see whats left.  Similary, if 2 and 6 have 1's in them try looking at blocks 5 and 3 for enough eliminated squares to ensure there is only one remaining spot for 1 in those blocks. Once you've tried all the possibilities for 1, move on to 2, and keep going until you've been through the whole list of numbers once without finding any new entries.\n\nFinally, a trick I really like, is when a block has only one row or one column with space remaining. You know that the remaining numbers in this mostly complete block will be in that row or column, and so you can use that information and apply it to other blocks even though you could not exactly place the numbers in the mostly complete block.\n\nHope that helps!
303259
n = N/(1+Ne(e))\n\nWhere:\nn = number of samples\nN = total population\ne = margin of error\n\nI could not find the squared key on my board so i just manipulated it, however it is the same.
113822
A 100 by 100 matrix has 100 eigenvectors. I would use some software to do this. This is too hard to do by hand. Buy the student version of MATLAB for example. You might be able to find some free software on the internet as well.
230284
You use sampling if they are fairly well spread out. Count just 1/4th the area of your sample and multiply by 4. So if you are looking at a 1" square sample you just count what's in a 1/4" square grid.