Math linear algebra question?
822074
..ll<3,2>ll  = sqrt( <3,2> . <3,2> )\n= sqrt( 4(3)(3) +7(2)(2) ) \n=sqrt( 36 +28) = sqrt ( 64)\n= 8
879937
ax^2 + bx+c = 0\nax^2 + bx = -c\nx^2+(b/a)x+ ... = -c/a + ...\nx^2+(b/a)x+ (b/2a)^2 = -c/a + (b/2a)^2\n{x - (b/2a)}^2 = -4ac/4(a^2) + b^2/4(a^2)\nx - (b/2a)}^2 = + or - root{(-4ac + b^2)/4(a^2)}\nx= {-b + or - sqrt (b^2-4ac)}/2a\n\nTh
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
957836
4x+(-5y)+(-6x)+3y\n4x+(-6x)+(-5y)+3y\n=-2x-2y
963452
(-5xy + 2xyz) + (xy - 3xyz) \n\nRight, the brackets are a distraction and are not needed, since there is a + in the middle, so remove them:\n\n-5xy + 2xyz+ xy - 3xyz\n\nNow collect like terms...\n\n-5xy + xy = -4xy    and    2xyz - 3xyz = -xyz\n\nPut these answers together and you get:\n\n-4xy - xyz\n\nVoila!\n\n(-5xy + 2xyz) + (xy - 3xyz) = -4xy - xyz
424681
substituiting your values you get (the values you gave are not very clear, can't make out whether positive or negative.\nAssuming all values are positive...)\ny-2 = 2(x-5)\ny-2 = 2x -10\n2x -y - 8 =0  is of the form ax + by + c =0
744496
A row of AB is of the form RB where R is a row of A. Now, if\nR1, R2, ...Rk are linearly dependent, there are constants c1,..ck\nwhere\nc1*R1+c2*R2+..ck*Rk=0.\nNow apply B on the right to get your result.
459985
Find the inverse of the matrix\n5 -3\n4  2\nA 2x2 matrix has a formula for the inverse.\nThe inverse of the matrix\na b\nc d\nis \nd/(ad-bc) -b/(ad-bc)\n-c/(ad-bc) a/(ad-bc)\nThus, for the above problem\na = 5, b = -3, c = 4 and d = 2\nad-bc = 5(2) - (-3)(4) = 10+12 = 22\nThe inverse is\n1/11  3/22\n-2/11 5/22\nMultiply this times the  matrix\n-2\n5\n= \n(1/11)(-2) + (3/22)(5)\n(-2/11)(-2) + (5/22)(5)\n= \n1/2\n3/2\nThus, x = 1/2 and y = 3/2 or (1/2,3/2)\nFor the second one the matrix is\n3 2 2\n1 2 -1\n2 -4 1\nThe inverse of this is\n1/14 5/14 3/14\n3/28 1/28 -5/28 \n2/7 -4/7  -1/7\nTake the inverse times the matirix\n3\n5\n0\nto get the matrix\n2\n1/2\n-2\nAnswer: x = 2, y = 1/2 and z =-2 or (2,1/2,-2)
1159782
you can use linear algebra, look up gauss jordan elimination. You will want to put each equation in its own row inside a matrix. Then you will have a coefficient matrix and an augmented matrix. Row reduce the left side and you will have your desired answers on the right. If you have a TI-83+ type in your matrix and hit RREF([A]) and it will do this for you. Go to the link I provided and scroll down a bit and it will show you.
798164
A = 3ax + 4by - 2ay - 6bx\n= 3x.(a - 2b) - 2y.(a - 2b)\n= (3x - 2y).(a - 2b)\n\nB = 2ax - 2by + ay - 4bx\n= 2x.(a - 2b) + y.(a - 2b)\n= (2x + y).(a - 2b)\n\nA/B = (3x - 2y).(a-2b) / [(2x + y).(a - 2b)]\n= (3x - 2y) / (2x + y)
738212
you need to set up a matrix equation.  \nthe rows should look like this:\n\n1      ,   -2    ,     1    ,      / -8\n3      ,   -3    ,    -2    ,     /-5\n-2     ,   -3     ,    3      ,    /5\n\nOnce you set this up follow the rules of matrices to get this to a unit matrix.  Once you have done this the three numbers on the right will become your answers for x, y and z.  x being which ever row has the 1 in the first column.  y being which ever row has the 1 in the second column.  z being which ever row has the 1 in the third column.
985452
Firstly, there is a formula. However, its quite easy to forget - so why not learn how to actually come up with it?\nSuppose you are given coordinates (a,b) in the first system. That really corresponds to the point a[2,1] + b[2,0]. Writing that in matrix form, we have: (// denoting next line)\n[2 2 // 1 0][a // b]\nNow, we want to express it as c[-2,-1]+d[0,1]. Writing that in matrix form we have:\n[-2 0 // -1 1][c // d].\nSo how do we work out [c // d]? Just multiply on the left by the inverse of that second matrix.\nSo it would be [-2 0 // -1 1]^(-1) * [2 2 // 1 0].\nThat turns out to be [-1 -1 // 0 -1].
199628
Assuming, of course, that each of the vectors has 3 entries...\n\nDo you know how to put a matrix formed from those four vectors in row-echelon form?\n\nI'm trying to decide the best way to teach this to you, which is why I need to know if you know how to get the matrices in REF, but the final "Answer" is to see if three of those vectors are linearly independent. If three of them are linearly independent, then they must span R^3.\n\nFor your second question, no. Remember that a singular matrix is a square matrix that is not invertible. Therefore, a singular matrix will not have a pivot in every column, which means that the columns are linearly dependent, which means that the rank is less than the dimension. Any square matrix with a column that is all 0's will will be singular.
1084623
| . x . x |\n| -x . -x |\n\nwhere x is any real number and x ≠ 0 (The last part is only needed so that A will be non-zero). The transpose also works.\n\nMultiply it out yourself. The result has nothing to do with triangular matrices, so I also suggest that you do not merely cite special characteristics of such matrices in your answer.
683485
It's easy.\nYou have 2 linear equation:\n3x - y = -7  -> equation  1\nx + y = -9   -> equation  2\n\nNote that in equation 2 you can rewrite as follow:\n  x = -9 - y \nBy simply moving the y to the other side.\nSo you know what x is in terms of y.\n\nIn equation 1 you have:\n      3x - y = -7  \nBut you know x now in terms of y and you can SUBSTITUTE to equation 1.\n\n    3(-9 - y) - y = -7\nSolving and expanding the multiplication you have\n    3*(-9) -3*y - y = -7\n    -27 - 4*y = -7\n         -4y  = -7 + 27\n           y  = -5\nSo you know y.\n\nIf you want to know x, you SUBSITUTE back to equation 2, since there you know x in terms of y.\nSo:\n    x = -9 - y \n    x = -9 - (-5)\n    x = -4\n\nChecking the answers by SUBSTITUTING back: \n    x = -4\n    y  = -5\n\n3x - y = -7  -> equation  1\n3(-4) - (-5) = -7  \n-12  + 5 = -7   (correct)\n\nx + y = -9   -> equation  2\n-4 + (-5) = -9 (correct)\n\nHope you understand the process :)
19948
a/bX + c = d\n\nfirst add -c to both sides\n=> a/bX + c-c = d-c\n=> a/bX = d-c\n\nNext time multiply b on both sides\n=> b * a/bX = (d-c) * b\n=> aX = b(d-c)\n\nAnd last divide both sides by a\n=> X = b(d-c)/a\n\nin this case:\n\nX = -4
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.