Using the indicated row operation to change the matrix . (how do you get it?)?
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
100903
If A is an n-by-n symmetric positive definite matrix with real entries, that means ((v transpose)Av) > 0 for nonzero real n-by-1 column vectors v.\n\nConsider the 3x3 case with A equal to\n\n[ a b c ]\n[ b d e ]\n[ c e f ]\n\nLet u, v, and w be the nonzero column vectors\n\n[ 1 ]\n[ 0 ]\n[ 0 ]\n\n[ 0 ]\n[ 1 ]\n[ 0 ]\n\n[ 0 ]\n[ 0 ]\n[ 1 ]\n\nrespectively.  Then\n\n((u transpose)Au) = a > 0\n((v transpose)Av) = d > 0\n((w transpose)Aw) = f > 0\n\nCan you see how to generalize this for other values of n besides three and/or for matrices and vectors with complex entries?
755982
Subscripts\nSubscripts are special postfix operators placed in square brackets ([ ]) after a matrix operand. Subscript operations have the general form \n\noperand[row,column] \nwhere \noperand \nis usually a matrix name, but it can also be an expression or literal. \n\n\nrow \nrefers to an expression, either scalar or vector, for selecting one or more rows from the operand. \ncolumn \nrefers to an expression, either scalar or vector, for selecting one or more columns from the operand. \nYou can use subscripts to \n\nrefer to a single element of a matrix \nrefer to an entire row or column of a matrix \nrefer to any submatrix contained within a matrix \nperform a reduction across rows or columns of a matrix \nIn expressions, subscripts have the same (high) precedence as the transpose postfix operator (). Note that when both row and column subscripts are used, they are separated by a comma. \n\nSelecting a Single Element\nYou can select a single element of a matrix in two ways. You can use two subscripts (row, column) to refer to its location, or you can use one subscript to look for the element down the rows. For example, referring to the coffee example used earlier, to find the element corresponding to the number of cups that Linda drank on Monday, you can use either of two statements. \nFirst, you can refer to the element by row and column location. In this case, you want the second row and first column. You can call this matrix c21. \n\n   >  print  coffee[rowname=names];\n\n           COFFEE\n           JENNY          4         2         2         3         2\n           LINDA          3         3         1         2         1\n           JIM            2         1         0         2         1\n           SAMUEL         5         4         4         3         4\n\n   >  c21=coffee[2,1];\n   >  print c21;\n\n                               C21\n                                 3\n\nYou can also look for the element down the rows. In this case, you refer to this element as the sixth element of COFFEE in row-major order. \n   >  c6=coffee[6];\n   >  print c6;\n\n                                C6\n                                 3\n\n\nSelecting a Row or Column\nTo refer to an entire row or column of a matrix, write the subscript with the row or column number, omitting the other subscript but not the comma. For example, to refer to the row of COFFEE that corresponds to Jim, you want the submatrix consisting of the third row and all columns: \n   >  jim=coffee[3,];\n   >  print jim;\n\n\n                JIM\n                  2         1         0         2         1\n\nIf you want the data for Friday, you know that the fifth column corresponds to Friday, so you want the submatrix consisting of the fifth column and all rows: \n   >  friday=coffee[,5];\n   >  print friday;\n\n                          FRIDAY\n                               2\n                               1\n                               1\n                               4\n\n\nSubmatrices\nYou refer to a submatrix by the specific rows and columns you want. Include within the brackets the rows you want, a comma, and the columns you want. For example, to create the submatrix of COFFEE consisting of the first and third rows and the second, third, and fifth columns, submit the following statements: \n   >  submat1=coffee[{1 3},{2 3 5}];\n   >  print submat1;\n\n                 SUBMAT1\n                       2         2         2\n                       1         0         1\n\nThe first vector, {1 3}, selects the rows, and the second vector, {2 3 5}, selects the columns. Alternately, you can create the vectors beforehand and supply their names as arguments. \n   >  rows={1 3};\n   >  cols={2 3 5};\n   >  submat1=coffee[rows,cols];\n\nYou can use index vectors generated by the index creation operator (:) in subscripts to refer to successive rows or columns. For example, to select the first three rows and last three colu
748718
There are 3 types of row operations:\n\n1. Each element of a row may be multiplied or divided by a number.\n2. Two rows may be interchanged\n3. A multiple of one row may be added to or subtracted from another row.\n\nIt sounds a lot more complicated than it is, so check out the website below for a great example.
802863
| A - I(lamda)| = 0 ==> eigenvalue ... \n\nlamda is an eigenvalue...\n\nsolve the matrix first
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.
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
840094
You use the dollar notation '$' For example if you write\n\n=sum($M$126/A3)\n\nand you cut and paste it will always keep it M126 (but the A3 will change). If you just want the Row to stay the same the M$126 or if you just want the column to stay the same $M126.\n\nNote you can also cycle through these options with the 'F4' key (i.e. if you get your M126 by clicking on that cell while entering your formula).
333863
Diagonalisation is the answer here.  Not sure how this didn't turn up in a web search, as I found plenty of resources for it, but the idea is that you find a matrix that diagonalises your initial matrix, and this allows powers to be done trivially.\n\nAssuming you're aiming to get a matrix A to the power n:\n\nFind P such that:\nP^-1 A P = D, where D is diagonal.  Then:\nA = (P D P^-1) and:\nA^n = (P D P^-1)(P D P^-1)(P D P^-1)....\n\nThe P^-1 P terms cancel and you're left with:\nA^n = P D^n P^-1\n\n... and this is much easier to calculate as D^n is trivial :)\n\nA good reference is in the sources.  Hope that helps!
972195
Easy, type $A$10+B8. The $ before A will anchor the column whereas the one before 10 will anchor the row.
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.
930026
wellz why not just help me with that matrix problem? :oP
772981
Without a calculator?  I guess you'd expand along the third row or the third column since they each have 3 zeros.  Say you use the third column.  You would have -5\n\n(negative because of the\n+ - + -\n- + - +\n+ - + -\n- + - + pattern)\n\n times the 3x3 determinant formed by crossing out the third column and the 4th row which is\n\n-7 0 3\n-5 -5 0\n0 -5 0\n\nWhich now could be expanded along the third column to be \n\n3 times the determinant formed by crossing out the third column and fist row:\n\n-5 -5 \n0 -5\n\nwhich is 25\n\nSo -5 (3)(25) = -375
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
957828
Just open it with Excel.  It will automatically populate the columns.\n-MM
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).