Number of submatrices of a matrix with m rows and n columns?
480863
There's something missing in that hint, there must still be an 'n' in there somewhere.\n\nTo work out the general formula:\n\nThe number of 2*2 matrices is (n-1)(m-1). The number of 2*3 is (n-1)(m-2). The number of 2*4 is (n-1)(m-3)... and so on until you reach (n-1)(1) = (n-1)(m-(m-1))\n\nSo the total number of 2 * j matrices is\nSigma(j = 1) { (n-1)(m-j) }\n\nThen the number of 3 * j matrices is\nSigma(j = 1) { (n-2)(m-j) }\n\nAnd so on, up to\nSigma(j = 1) { (1)(m-j) }\n\nThe sum of all these gives the number of i * j matrices as:\n\nSigma(i=1 to n-1)Sigma(j=1 to m-1) {(n-i)(m-j)}
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by the number of sides and angles
320882
1 2 3 4 5 6 7 8\n<---[======)--->
3477
64 squares of 1x1, 49 squares of 2x2, 36 squares of 3x3, 25 squares of 4x4, 16 of 5x5, 9 of 6x6, 4 of 7x7, 1 of 8x8 - 204, at least [in case I missed something :-) ]
756126
you need to do the following:\nT( (1,-1))= (0, 2) = -6 (1,-1) - 2 (-3,2)  so the first column of your matrix is:\n-6\n-2\n\nnow\nT ( (-3,2) ) = ( -1, -5) = 17 (1,-1) + 6 (-3,2), so the second column of your matrix is:\n17\n6\n\nwhen you put them together you get:\n-6 17\n-2 6\n'
414270
Add the two tops and add two bottoms.Then the sum is\n(Top)[1+0 , 3+0] and (Bottom)[2+0 , 6+0] = E\nE is any 2x2 matrix and O is the 2x2 zeromatrix.\nAlways is E+O = O+E = E.\nThis is also true for nxn matrices.
1035025
A matrix is a rectangular array of numbers.  For our purposes, let z{m,n} stand for the number z in the mth row and nth column of a matrix.\n\nExample -- Consider the 2x2 matrix named ' A'.  Let A be composed of numbers z{1,1}=a,  z{1,2}=b,  z{2,1}=c, and z{2,2}=d.  \nOr more simply,   the matrix A is set up as\n| a   b |\n| c   d |\n\nBy definition:  If 'A' is square matrix, and if a matrix 'B' can be found such that  AB=BA=I, then A is said to be invertible, and B is called an inverse of A.\n\nNow, go back to our Example 2x2 matrix named 'A'.  If (ad - bc) is not equal to zero, then the inverse of A, called A^(-1), is equal to\n\n(1/ (ad-bc) ) times the matrix\n  |  d   (-b) |\n  | (-c)   a  |\nwhere z{1,1}=d,  z{1,2}= -b,  z{2,1}= -c, and z{2,2}=a.  \n\nOR\n\n |  (d/(ad-bc))     (-b/(ad-bc)) |\n |  (-c/(ad-bc))    (a/(ad-bc))  |\nwhere z{1,1} =(d/(ad-bc)),  z{1,2} =(-b/(ad-bc)),  z{2,1}=(-c/(ad-bc)), and z{2,2} =(a/(ad-bc)).
748718
There are 3 types of row operations:\n\n1. Each element of a row may be multiplied or divided by a number.\n2. Two rows may be interchanged\n3. A multiple of one row may be added to or subtracted from another row.\n\nIt sounds a lot more complicated than it is, so check out the website below for a great example.
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I dont know how to put this up here. I tried to put it in a nice format, but it would accept.\nBut you put the following in one matrix \n-2  6\n+2  2\n\nThen you put a * (multiply)\n\nThen you put the following in one more matrix\n\nx\ny\n\n\nThen put an ' = ' symbol\n\nThen put the following in a matrix\n\n40\n8
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.
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
94604
A quotation from\nhttp://www.ams.org/featurecolumn/archive/sylvester3.html\n\n"In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). "
725218
Modified Gram-Schmidt algorithm\n------------------- -----------------------\nA  is m x n,  with m ≥ n  s.t. rank(A) = n\n\n(a) For i = 1, ... , n \n`````````w_i = a_i\nEnd (a) For\n(b) For i = 1, ... , n\n`````````r_ii = ||w_i||\n`````````q_i = (w_i)/(r_ii)\n`````````(c) For j = i + 1, ... , n\n``````````````````r_ij = <q_i, w_j>\n``````````````````w_j = w_j - (r_ij)(q_i)\n`````````End (c) For\nEnd (b) For
356990
Gauss Jordan method for this:\n\n| 6 -5 | 4 |\n| -6 5 | -2 |\n\nR1/6 -> R1\n| 1 -5/6 | 2/3 |\n| -6 5 | -2 |\n\n6R1 + R2 -> R2\n| 1 -5/6 | 2/3 |\n| 0 0 | 2 |\n\nFrom this, we see that there is no solution, since we have a row of 0's on the left of the matrix.\n\nIf you try to solve this by substitution or by any other method you know, you'll find that you can't.  It turns out that these are parallel lines, and parallel lines don't intersect, so you won't get an answer for this particular set of equations.
570250
7.8\ntry www.matweb.com, a great site for more details.
909215
It can be encoded with another 1*2 matrix through addition.\n\nFor example,\n\noriginal matrix = (x   y)\n\nencoding matrix = (a b)\n\nCoded matrix = (x+a    y+b)
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sample space {1,2,3,4,5,6}