a matrix contains 60 elements. which of the following cannot equal the number of rows of the matrix?
964841
Figure it out - when you divide 60 by each of those numbers, which one leaves a remainder?
596773
Matrix means rows and columns. There lots of problems based on Matrix in mathematics like Determinants. \n\nThe movie matrix is sci-fi movie. Its not real.
1087743
Matrix - the no. of rows and columns are not necessarily equal.\nDeterminants - they r equal.\nMatrix - elements are enclosed in (), { }, or [ ]\nDeterminants - elements r shown between two vertical lines. I I\nMatrix - it cant hav a value. its just an arrangement.\nDeterminant - has a value.\nMatrix - if all elements r multiplied by a constant the matrix is multiplied to the constant.\nDeterminant - if each of any row or column is multiplied by a constant, the determinant is mulitplied to the constant.\nPhew * Need some lemonade...
283770
This is about Ramsy numbers , R(17,3)=3.\nThis is a famous problem, but the proff is not easy.
783847
Assume that the numbers of elements of A1, A2, ..., An are m1, m2, ..., mn, respectively. Then their union has at most m1+m2+...+mn elements (exactly so many if the sets are disjoint, but if they have common elements then their union has less then m1+m2+...+mn elements), which is finite, because it is the sum of finitely many finite numbers.
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)}
702805
B...multiply each number by 4 in the matrix
551932
Assuming your matrix is of order nxn and you want submatrix of order mxm.\n\nThen there is (n-m+1)^2 submatrices.\n\nSo you can run this loop(Java code)\n\nfor(i=0;i==n-m;i++)\n{\nint k;\nfor(k=0;k==n-m;k++)\n{\nint p,q;\nfor(p=0;p<m;p++)\n{\nfor(q=0;q<m;q++)\n{\nprint("  ");\nprint(a[p][q]);\n}\nprintln();\n}\n}\n\n}\n\nTo print all the submatrices iterate over this procedure canging m from 1 to n.
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.
195373
Try this:\n\n#include <stdio.h>\n\nmain()\n{\nint size,array[50],n,m,border,max,buf,step;\nint i, j, temp;\n\nprintf("Input number of elements in array:");\nscanf("%d",&size);\n\nfor(n=0;n<size;n++) {\n  printf("input %d. element of array:",n+1);\n  scanf("%d",&array[n]);\n  }\n\n for (i=(size-1);i>=0; i--)\n  {\n   for(j=1;j<=i;j++)\n    {\n     if(array[j-1]>array[j])\n      {\n       temp=array[j-1];\n       array[j-1]=array[j];\n       array[j]=temp;\n      }\n     }\n   }\n\nfor(n=0;n<size;n++)\n   printf("%d. element: %d&#92;n",n+1,array[n]);\n\n}\n\n\n\nHERE IS MY OUTPUT:\n\nInput number of elements in array:8\ninput 1. element of array:6\ninput 2. element of array:2\ninput 3. element of array:3\ninput 4. element of array:5\ninput 5. element of array:1\ninput 6. element of array:4\ninput 7. element of array:5\ninput 8. element of array:7\n1. element: 1\n2. element: 2\n3. element: 3\n4. element: 4\n5. element: 5\n6. element: 5\n7. element: 6\n8. element: 7
1102914
10\nThey are factors of 60 (in order from largest to smallest).
50989
Eleven if you count the two adjacent sides of the polygon, otherwise nine.\n\nThinking outside the box answer:\nUnlimited if one does not limit themselves to the other vertices as termini.
947340
a) 72*60*24*30=3110400\nb) 3110400*12 = 37324800\nc) 37324800*72=...................................... :))
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.
515192
If it's a square, (60 x 60) then probably 8.\nIf it's a circle (r=60) then probably 10.\nIf it's a cirlce (d=60) then probably 6
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.
836747
2,3,1 (nothing goes there) ,6\n\nHope it helps :]