Row Operations for Matrices?
955196
is like using the equations.  to eliminate x multiply the first equation by -1 and add it to the second equation:\n\nx+y=8\n-2y = -4\n\nThen use the second equation to eliminate y from the first:\nx = 6\n-2y=-4\n\nie, \nx=6\ny=2
717242
I apologize for the notation in advance.\n\nCholesky factorization assumes that a matrix is symmetric and positive definite.  Symmetric is easy to determine.  Since Cholesky is fairly efficient, it can be used to determine if in fact a symmetric matrix is positive definite.  \n\nThe gist of the algorithm is this:  let's assume that we can write a symmetric positive definite matrix A as\nA=U'*U (' means transpose), where U is upper triangular.  To think about this, let's write the matrix A in blocks:\n\nA=(a_11 A_1,A'_1 A_22)\n\nwhere a_11 is the 1,1 element of A, A_1 is the remainder of row 1 of A, and A_22 is the 2,2 cofactor of A.  Similarly write U as \n\nU=(u_11 U_1,0 U_22)\n\nThen U'*U=A gives us\n\nu_11*u_11=a_11\nu_11*U_1=A_1\nU'_1*U_1 + U'_22*U_22 = A_22\n\nThis doesn't look like it gets us that far, but notice that U'_22*U_22 is a symmetric positive definite matrix, to which the preceding steps of the algorithm can be applied.  That's the way Cholesky works; proceed as above until done.  \n\nQuick example:\nA=(1 2 3,2 8 8,3 8 11)\nThen u_11=sqrt(1)=1\nU_1=(1/u_11)*A_1=(1/1)*(2 3)=(2 3) (this is a row vector)\n\nU'_22*U_22 = A_22 - (U'_1*U_1)\n=(8 8,8 11)-(2 3)'*(2 3)=(4 2,2 2)\n\nRepeat with this new matrix:\nu_22=sqrt(4)=2\nU_2=(1/2)*(2)=(1) (writing as a row vector to emphasize similarity with above)\nU_'33*U_33 = A_33 - U'_2*U_2 = 2-1*1=1\n\nFinally, u_33=sqrt(1)=1 and the algorithm terminates.  So your matrix U is\n\nU=(1 2 3,0 2 1,0 0 1)\n\nIf at any point, your proposed value for u_jj is the square root of a negative number, the matrix you started with is not positive definite.
595375
divide the first row by 2\n1 1/2\n1  -1\n\nSubtract the first row from the second row:\n1  1/2\n0  -3/2\n\nMutiply the second row by -2/3\n1  1/2\n0   1\n\nSubtract 1/2 of the second row from the first row:\n1  0\n0  1\n\nAnd you have the identity matrix
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.
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).
470202
Rank is basically the number of nonzero rows in a row-echelon reduced matrix.  So if you have a square matrix with one row of zeroes, you may be looking at something like this:\n\n| 1 0 0 |\n| 0 1 4 |\n| 0 0 0 |\n\nadj(A) is found by finding the matrix made up of the cofactors of A, and then transposing it, and changing signs in a checkerboard pattern in the matrix.  Creating the cofactor matrix is easy - for the new matrix C, Cij is found by finding the determinant of the matrix that results if you elimiate row i, column j in matrix A.  The cofactor matrix for the above example would be\n\n| 0 0 0 |\n| 0 0 0 |\n| 0 4 1 |\n\nand its transpose would be\n\n| 0 0 0 |\n| 0 0 -4 |\n| 0 0 1 |\n\nwhich is a matrix of rank 1.\n\nThe row of zeros creates multiple rows of zeros when creating the cofactor matrix.  That makes sense - when finding the determinant of a matrix with a row of zeros, the determinant always equals 0.  It turns out that the only time that you create a nonzero row in your cofactor matrix is when you eliminate the single row of zeros.
1126894
This normally means that you are going to be using Gaussian (or Gauss-Jordan) reduction, where ther eare three "moves" that are allowed.\n\n1) You can switch any two rows in the matrix.\n2) You can multiply any row in the matrix by a non-zero scalar.\n3) You can add a multiple of one row to another.\n\nUsing these three "moves," the goal is to reduce the matrix into a usable form (the echelon form, for example).
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!!!
201316
This website may help you.  Good luck.\n\nhttp://www.sosmath.com/matrix/matrix.html
1183592
If your matrix is not square, then you can compute a generalized inverse matrix.  There are some links below.  There are many generalized inverses for matrices (other than those that are square and invertible; those only have one generalized inverse, which is its inverse).  But there is only one Moore-Penrose inverse for any given matrix.\n\nedit: reetz, I have already reported you for copying.  I suggest you back off.
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
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.
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)).
346932
well, in case I don't answer sufficiently, I provided a link for Wikipedia to define the ranks of matrices.\n\nBasically, ranks are when the matrices are reduced and you want to know the number of independent rows and columns.\n\nSince A^2 = I, this confirms A is nxn, so we're off to a good start.\n\nThe product of each row i x column i = 1, and row i x column j = 0, when i <> j.\n\nI think rank(I+A) + rank(I-A) somewhere is the solution by comparing the number of independent rows left of I+A and noticing there's only so many independent rows left in I-A.\n\nOne might wonder what does rank(I+A) + rank(I-A) suggest about the equation... (I+A)x(I-A)\n\n(I+A)x(I-A) = matrix[I(i,j)+A(i,j)] x matrix[I(k,l)-A(k,l)]\n            = matrix[sigma(row(i)xcolum(l))]\n            = matrix[sigma(j)(I(i,j)*I(i,l)-I(i,j)*A(j,l)+I(i,j)*A(j,l) -A(i,j)*A(j,l))]\n            = matrix[sigma(j)(I(i,j)*I(i,l)-A(i,j)*A(j,l))]\n            = I^2-A^2\n            = I - I\n            = 0\n\nSorry I couldn't answer your question completely.
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.
21539
If you have a system of equations with two variables of the form ax + by = c, dx + ey = f, where a,b,c,d,e,f are the coefficients, and x and y are variables, you can set up the equation as \n\n[a b][x] = [c]\n[d e][y] = [f]\n\n(ignore the fact there are two equals signs there, they're only there for formatting reasons)\n\nwhere the square brackets are a crude attempt on by behalf to 'draw' matrices and column vectors.\n\nTo solve the system you need to find the inverse of the matrix of the right hand side (assuming it's invertible to begin with), and "right multiply" (that is multiply from the right of each side of the equation) both sides of the equation.\n\nThat should give you:\n\n[1 0][x] = [ inverse of matrix ][c]\n[0 1][y] = [ ........................ ][f] \n\nThen it's just a matter of performing the calculation.\n\n(sorry for the crappy formatting of the post)
222586
um for a 2x2 matrix, there is a general formula for the matrix. which is\n\n1/ad-bc  *  [d,   -b]      if the matrix is   [a, b]\n............[-c,   a]      ...................[c, d]\n\nso then the answer would be\n\n1/(1)(6) - (6)(3)   *   [6,   -6]\n........................[-3,  1 ]\n\n[-1/2,  1/2  ]\n[ 1/4,  -1/12]\n\nhope this helps. elt me know if u need more help