E+O=???  (this is a matrix equation)?
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.
44938
Let \nBlue = x\nYellow = y\n\nThen:\n2x + y = Green   [Given]
926227
Well Aq = b. where A is the matrix of coefficients, q is [x; y; z; w], and b is the answer [-2; 2; -11; -3] So we are looking for the Rx = d, that corresponds to the original equation. The aumented matrix is \n     5     3     1     5    -2\n     3     2     2     4     2\n     2     4     3    -3   -11\n     4    -3    -2     2    -3\nThe first step is to create a 1 in 1,1 so we divide Row1 by 5.\nThen subtract 3xRow1 from row 2, 2xRow1 from Row3, 4xRow1 from Row4\nThis is the result\n         1    0.6     0.2    1       -0.4\n         0    0.2     1.4    1        3.2\n         0    2.8     2.6   -5    -10\n         0   -5.4    -2.8   -2      -1.4\nContinue down and to the right with row operations in the same vein until the matrix is in row echelon form then go up and to the left in order to create 0s in all the non-step positions and get the matrix in reduced row echelon form.\n\nThe reduced row echlon form of  the augmented matrix [A b] is \n     1     0     0     0    -2\n     0     1     0     0    -1\n     0     0     1     0     1\n     0     0     0     1     2\n(Heh look at the nice numbers. You got this out of a textbook didn't ya?)\nThis was of course gotten by the Gauss-Jordan Method.\nSo now it's just a matter of reading off the coefficients\n1q1 + 0q2 + 0q3 + 0q4 = -2\n0q1 + 1q2 + 0q3 + 0q4 = -1\n0q1 + 0q2 + 1q3 + 0q4 =  1\n0q1 + 0q2 + 0q3 + 1q4 =  2\n\nSo q = [-2; -1; 1; 2] and x = -2, y = -1, z = 1, w =2.\n\nIf you want all the row operations, well there are an awful lot of them and they're all simple and straightforward. It's simply a matter of carefully working through the algorithm, so you are better off working them out for yourself. You'll probably get them before I can type them all in.
437832
Homework.\n\nSuppose a1*x+a2*Tx+..+am*T^(m-1)x=0.\n\nApply T^(m-1) to get a1=0.\nApply T^(m-2) to get a2=0, etc.
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
727014
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
550918
Type "=B1-C1+D1*A1" without the quotes inside another cell such as E1.  This is assuming that your equation is the correct algebraic form and solved for one variable.  Your text is a bit jumbled so I didn't parse all of it.
542644
Here are a few websites which explain it fairly well...\n\nhttp://www.mai.liu.se/~halun/matrix/\n\nhttp://www.easycalculation.com/matrix/matrix-multiplication.php\n\nThat one has a matrix calculator online. Good luck!
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
827717
I presume that's a pair of 2x2 matrices?  The answer is B.  I hate just giving the answer, but I don't see any reasonable way to discuss matrix multiplication in this format.  Please be sure you understand WHY it's B before you turn it in.
219213
So you need to find a lower diagonal matrix L and an upper diagonal matrix U such that L*U equals the given matrix. The most straightforward way of doing this is to perform Gaussian elimination while keeping track of all the row operations you perform. The row operations are used to construct the matrix L, and the end product of the Gaussian elimination will be U.\n\nStart with the 1 in the top left corner and use it to eliminate all the non-zero terms below it. So take twice the first row and subtract it from the second row, giving you:\n1 1 1\n0 0 0\n3 3 3\n\nWhile you are doing this you should keep track of a matrix that will undo all the operations you have done (this will be your L matrix in the end). Start with the identity matrix and add a 2 in the first column of the second row (it's a +2 since we subtracted twice the first row but we are looking for a matrix that will undo all the operations we are performing and get us back to the original matrix)\n1 0 0\n2 1 0\n0 0 1\n\nNow eliminate the 3 in the bottom left corner by taking the third row and subtracting 3 times the first row, giving you:\n\n1 1 1\n0 0 0\n0 0 0\nwhich is upper triangular (so it'll be your U) and an L matrix:\n1 0 0\n2 1 0\n3 0 1\nwhich is lower triangular. You can confirm that L*U does in fact equal your original matrix, so you're done.
732131
ha! I did this today in Algebra!!\n\nOkay, basically you take each of the coordinates and plug them into this fomula\n\n\nplusorminus 1/2  det| x1 y1 1|\n                                 | x2 y2 1|\n                                 | x3 y3 1|         solve for 1/2 of this detemanant just like you would solve for a 3 x 3 matrix.  You answer of course must always be positive.\n\nfor some reason, the two lines under plusorminus... got moved over, they should fall directly under |x1 y1 1|
710964
ab=6\n\nit is an equation because of =
838286
well Honestly dear Megan i don't think its correct , coz equation should have ONE " = " \n\nit must be something like this \n-4r ² ± 21r=r+13\n\nGood Luck
241815
If this matrix is A, we want to solve an equation AX=B, to solve this equation we have to cancel A, so multiply both sides by an inverse, A^(-1) and obtain X=BA^(-1).\nIf the matrix in non invertible, we cannot cancel it as easily. At least I assume that was the cancellation you requested.