Write the augmented matrix for this system . But do not solve?
398921
To make an augmented matrix, just take your coefficients and put them in a matrix, then draw a vertical line and place the constants on the right side, like this:\n\n| -2 0 9 | 2 |\n| 0 -2 9 | 22 |\n| 3 5 -2 | 10 |\n\nFill in 0's for any variables that aren't there (i.e. there's no y in the first equation, so there's a 0 in its place in the matrix).
696922
Let's say it's like this:\n[1 2 3]\n[4 5 6]   } Pretend the whole thing is surrounded by two big\n[7 8 9]     brackets\n\n[9 8 7]\n[6 5 4] And that the top is multiplied by this\n[3 2 1]\n\n[1 2 3]------[9 8 7]   (I'm using---as spaces)\n[4 5 6]--X--[6 5 4] \n[7 8 9]------[3 2 1]\n Multiply the first number in row one of the 1st matrix by the first column of the 2nd matrix ( 9 6 3 ). Then multiply the second number in row one by the second column ( 8 5 2 ) and then 3rd number by the 3rd column ( 7 4 1 ). When you're done, you should have the first row of a newer giant matrix set up...\n\n[1(9) + 2(6) + 3(3)---1(8) + 2(5) + 3(2)---1(7) + 2(4) + 3(1)]\n\nNow, take the 2nd row of of the first matrix and multiply it the same way...(now you have the first and second rows I showed you)\n\n[1(9) + 2(6) + 3(3)---1(8) + 2(5) + 3(2)---1(7) + 2(4) + 3(1)]\n[4(9) + 5(6) + 6(3)---4(8) + 5(5) + 6(2)---4(7) + 5(4) + 6(1)]\n\nDo you notice a pattern? The column numbers that are multiplied are lined up right on top of each other... this will save you time in the future. The row numbers change, but you know they are always in the same order.\nNow, multiply the last row of the first matrix by the columns of the second matrix...(here is how the final matrix looks)\n\n[1(9) + 2(6) + 3(3)---1(8) + 2(5) + 3(2)---1(7) + 2(4) + 3(1)]\n[4(9) + 5(6) + 6(3)---4(8) + 5(5) + 6(2)---4(7) + 5(4) + 6(1)]\n[7(9) + 8(6) + 9(3)---7(8) + 8(5) + 9(2)---7(7) + 8(4) + 9(1)]\n\nNow, multiply the numbers...\n[9 + 12 + 9---8 + 10 + 6---7 + 8 + 3]\n[36 + 30 + 18---24 + 25 + 12---28 + 20 + 6]\n[63 + 48 + 27---56 + 40 + 18---49 + 32 + 9]\n\nAdd up the numbers to get the final resulting matrix...\n[30  24  18]\n[74  61  54]\n[111 114  90] \nP.S. Sorry I didn't use your numbers... this is the answer to the matrix I made up, but at least it shows you how to do it right? Hope this helps!!!
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)}
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
308173
You haven't explained the question properly! These matrix can be formed any way!
664485
nope \n5x+5y\n\nthe 4y and -4y cancel each other out
458991
No, but if you go to List mode you can enter your matrix as a list and manipulate it.  Check out the owner's manual for help on how to do it.  I've included a link to it.  Hope this helps.
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
753930
The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
469735
The system of equations:\n[a11 a12 a13; a21 a22 a23]\n[1 -3 4; 2 3 1]\n\nPerforming indicated row operation:\n[a11 a12 a13; a21' a22' a23']\n[1 -3 4; -1 -6 3]\n\na21' = a11 - a21 = 1 - 2 = -1\na22' = a12 - a22 = -3 - 3 = -6\na23' = a13 - a23 = 4 - 1 = 3
236017
sorry i don't know mathematica but you can solve it easily using matlab. Enter rthe coefficient matrix as follows for example,\nA=[1 2 3 4; 2 3 3 4;........]\nand the right hand side\nb=[ 2 3 4 4]\nonly for illustration. ";"indicates columns.\nA/b solves this system and gives the solution vector x.
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.
148434
So, continuing with mr_w's excellent example, we have 4 equations in 4 unknowns. So you can set this up as a matrix-vector problem.  Define matrix M such that M * Number of games bought vector = Cost & total games vector.  Assigning the rows of M in order to the total cost, the PS3 and Xbox360 cost, the PS1 and Xbox360 cost, and the total number of games, \nM = \n70 150 300 450\n0 0 300 450\n70 0 0 450\n1 1 1 1\nNumber of games bought vector =\na\nb\nc\nd\nand Cost & total games vector =\n13910 12000 3050 54\nThe solution will find the values of a, b, c and d.\nWhen I ran the solution I ended up with noninteger values for each and the values checked with the problem statements.  I hope those values were mr_w's solution.
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
737238
the 3 x 3 matrix A times the 3 x 3 matix B will yield a 3 x 3 matrix AB\n\nAB1,1 = (A1,1)(B1,1) + (A1,2)(B2,1) + (A1,3)(B3,1)\nAB1,1 = (-3)(-1) + (6)(0) + (2)(-5) = -7\n\nAB1,2 = (A1,1)(B1,2) + (A1,2)(B2,2) + (A1,3)(B3,2)\nAB1,2 = (-3)(8) + (6)(-3) + (2)(0) = -42\n\nAB2,1 = (A2,1)(B1,1) + (A2,2)(B2,1) + (A2,3)(B3,1)\nAB2,1 = (-4)(-1) + (8)(0) + (1)(-5) = -1\n\netc.
118612
Kerplode is right - but that's kind of complicated.  Basically you solve both in exactly the same way.  The only difference is that instead of an = you have an > or <.  \n\nIt just looks different, there really is no difference in how you solve them.
201908
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2