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Gauss-Jordan Elimination in .NET Connect pdf417 2d barcode in .NET Gauss-Jordan Elimination




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2.1 Gauss-Jordan Elimination using barcode drawer for none control to generate, create none image in none applications. GS1 Barcodes Knowledge Elimination on Column-Augmented Matrices Consider the linear matrix equation a12 a22 a32 a42 a13 a23 a33 a43 a14 x11 a24 x21 a34 x31 a44 x41 x12 x22 x32 x42 a11 a21 a31 a41 x13 x23 x33 x43 y11 y21 y31 y41 0 1 0 0 0 0 1 0 y12 y22 y32 y42 y13 y23 y33 y43 y14 y24 y34 y44 =. b11 b21 b31 b41 b12 b22 b32 b42 b13 b23 b33 b43 1 0 0 0. 0 0 0 1. (2.1.1).

Here the raised do none none t ( ) signi es matrix multiplication, while the operator just signi es column augmentation, that is, removing the abutting parentheses and making a wider matrix out of the operands of the operator. It should not take you long to write out equation (2.1.

1) and to see that it simply states that xij is the ith component (i = 1, 2, 3, 4) of the vector solution of the jth right-hand side (j = 1, 2, 3), the one whose coef cients are bij , i = 1, 2, 3, 4; and that the matrix of unknown coef cients yij is the inverse matrix of aij . In other words, the matrix solution of [A] [x1 x2 x3 Y] = [b1 b2 b3 1] (2.1.

2). where A and Y are square matrices, the bi s and xi s are column vectors, and 1 is the identity matrix, simultaneously solves the linear sets A x1 = b 1 and A Y = 1 (2.1.4) Now it is also elementary to verify the following facts about (2.

1.1): Interchanging any two rows of A and the corresponding rows of the b s and of 1, does not change (or scramble in any way) the solution x s and Y. Rather, it just corresponds to writing the same set of linear equations in a different order.

Likewise, the solution set is unchanged and in no way scrambled if we replace any row in A by a linear combination of itself and any other row, as long as we do the same linear combination of the rows of the b s and 1 (which then is no longer the identity matrix, of course). Interchanging any two columns of A gives the same solution set only if we simultaneously interchange corresponding rows of the x s and of Y. In other words, this interchange scrambles the order of the rows in the solution.

If we do this, we will need to unscramble the solution by restoring the rows to their original order. Gauss-Jordan elimination uses one or more of the above operations to reduce the matrix A to the identity matrix. When this is accomplished, the right-hand side becomes the solution set, as one sees instantly from (2.

1.2). A x2 = b2 A x3 = b3 (2.

1.3). 2. . Solution of Linear Algebraic Equations Pivoting In Gauss-Jordan e limination with no pivoting, only the second operation in the above list is used. The rst row is divided by the element a11 (this being a trivial linear combination of the rst row with any other row zero coef cient for the other row). Then the right amount of the rst row is subtracted from each other row to make all the remaining ai1 s zero.

The rst column of A now agrees with the identity matrix. We move to the second column and divide the second row by a22 , then subtract the right amount of the second row from rows 1, 3, and 4, so as to make their entries in the second column zero. The second column is now reduced to the identity form.

And so on for the third and fourth columns. As we do these operations to A, we of course also do the corresponding operations to the b s and to 1 (which by now no longer resembles the identity matrix in any way!). Obviously we will run into trouble if we ever encounter a zero element on the (then current) diagonal when we are going to divide by the diagonal element.

(The element that we divide by, incidentally, is called the pivot element or pivot.) Not so obvious, but true, is the fact that Gauss-Jordan elimination with no pivoting (no use of the rst or third procedures in the above list) is numerically unstable in the presence of any roundoff error, even when a zero pivot is not encountered. You must never do Gauss-Jordan elimination (or Gaussian elimination, see below) without pivoting! So what is this magic pivoting Nothing more than interchanging rows (partial pivoting) or rows and columns (full pivoting), so as to put a particularly desirable element in the diagonal position from which the pivot is about to be selected.

Since we don t want to mess up the part of the identity matrix that we have already built up, we can choose among elements that are both (i) on rows below (or on) the one that is about to be normalized, and also (ii) on columns to the right (or on) the column we are about to eliminate. Partial pivoting is easier than full pivoting, because we don t have to keep track of the permutation of the solution vector. Partial pivoting makes available as pivots only the elements already in the correct column.

It turns out that partial pivoting is almost as good as full pivoting, in a sense that can be made mathematically precise, but which need not concern us here (for discussion and references, see [1]). To show you both variants, we do full pivoting in the routine in this section, partial pivoting in 2.3.

We have to state how to recognize a particularly desirable pivot when we see one. The answer to this is not completely known theoretically. It is known, both theoretically and in practice, that simply picking the largest (in magnitude) available element as the pivot is a very good choice.

A curiosity of this procedure, however, is that the choice of pivot will depend on the original scaling of the equations. If we take the third linear equation in our original set and multiply it by a factor of a million, it is almost guaranteed that it will contribute the rst pivot; yet the underlying solution of the equations is not changed by this multiplication! One therefore sometimes sees routines which choose as pivot that element which would have been largest if the original equations had all been scaled to have their largest coef cient normalized to unity. This is called implicit pivoting.

There is some extra bookkeeping to keep track of the scale factors by which the rows would have been multiplied. (The routines in 2.3 include implicit pivoting, but the routine in this section does not.

) Finally, let us consider the storage requirements of the method. With a little re ection you will see that at every stage of the algorithm, either an element of A is.
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