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The left 3 3 block is now upper triangular. in .NET Printer Code 128A in .NET The left 3 3 block is now upper triangular.

The left 3 3 block is now upper triangular. generate, create code 128c none for .net projects Use Mobile Phone to Scan 1D and 2D Barcodes Appendix I. Review of Linear Algebra Next, we have to wipe ou t, one by one, the elements in this block above the diagonal. Thus, subtract row 3 from row 1: 1 2 0 0 1 0 0 0 1 Tableau 6 3/4 1/2 1/6 0 1/4 1/2 0 1/3 . 0.

Finally, subtract two ti code128b for .NET mes row 2 from row 1: I 1 0 0 0 1 0 0 0 1 Tableau 7 A 1 5/12 1/2 2/3 1/6 0 1/3 . 1/4 1/2 0.

This is the nal tableau . The last three columns now form A 1 . Once you have calculated A 1 , you can solve the linear system Ax = b by computing x = A 1 b.

However, you can also incorporate the latter in the Gauss Jordan iteration, as follows. Again let A be the matrix in (I.22), and let, for example, 1 b = 1 .

1 Insert this vector in Tableau 1: Tableau 1 A b 2 4 2 1 1 2 3 1 1 1 1 1 I 1 0 0 1 0 0. 0 0. 1. and perform the same row operations as before. Then Tableau 7 becomes I 1 0 0 0 1 0 0 0 1 Tableau 7 A 1 b A 1 5/12 5/12 1/2 2/3 1/2 1/6 0 1/3 . 1/4 1/4 1/2 0.

This is how matrices wer visual .net Code 128 Code Set B e inverted and systems of linear equations were solved fty and more years ago using only mechanical calculators. Nowadays of course you would use a computer, but the Gauss Jordan method is still handy and not too time consuming for small matrices like the one in this example.

. The Mathematical and Statistical Foundations of Econometrics I.7. Gaussian Eliminatio barcode standards 128 for .

NET n of a Nonsquare Matrix The Gaussian elimination of a nonsquare matrix is similar to the square case except that in the nal result the upper-triangular matrix now becomes an echelon matrix: Definition I.10: An m n matrix U is an echelon matrix if, for i = 2, . .

. , m, the rst nonzero element of row i is farther to the right than the rst nonzero element of the previous row i 1. For example, the matrix 2 0 1 0 U = 0 0 3 1 0 0 0 4 is an echelon matrix, and so is 2 0 1 0 U = 0 0 0 1 .

0 0 0 0 Theorem I.8 can now be generalized to Theorem I.11: For each matrix A there exists a permutation matrix P, possibly equal to the unit matrix I, a lower-triangular matrix L with diagonal elements all equal to 1, and an echelon matrix U such that PA = LU.

If A is a square matrix, then U is an upper-triangular matrix. Moreover, in that case PA = LDU, where now U is an upper-triangular matrix with diagonal elements all equal to 1 and D is a diagonal matrix.8 Again, I will only prove the general part of this theorem by examples.

The parts for square matrices follow trivially from the general case. First, let 2 4 2 1 (I.35) A = 1 2 3 1 , 1 1 1 0.

Note that the diagonal e lements of D are the diagonal elements of the former uppertriangular matrix U.. Appendix I. Review of Linear Algebra which is the matrix (I.2 2) augmented with an additional column. Then it follows from (I.

31) that 1 0 0 2 4 2 1 P2,3 E 3,1 (1/2)E 2,1 ( 1/2)A = 0.5 0 1 1 2 3 1 0.5 1 0 1 1 1 0 2 4 2 1 = 0 3 0 1/2 = U, 0 0 2 1/2 where U is now an echelon matrix.

As another example, take the transpose of the matrix A in (I.35): 2 1 1 4 2 1 AT = 2 3 1 . 1 1 0 Then P2,3 E 4,2 ( 1/6)E 4,3 (1/4)E 2,1 ( 2)E 3,1 ( 1)E 4,1 ( 1/2)AT 2 1 1 0 2 0 = 0 0 3 = U, 0 0 0 where again U is an echelon matrix.

I.8. Subspaces Spanned by the Columns and Rows of a Matrix The result in Theorem I.

9 also reads as follows: A = BU, where B = P 1 L is a nonsingular matrix. Moreover, note that the size of U is the same as the size of A, that is, if A is an m n matrix, then so is U . If we denote the columns of U by u 1 , .

. . , u n , it follows therefore that the columns a1 , .

. . , an of A are equal to Bu1 , .

. . , Bun , respectively.

This suggests that the subspace spanned by the columns of A has the same dimension as the subspace spanned by the columns of U . To prove this conjecture, let V A be the subspace spanned by the columns of A and let VU be the subspace spanned by the columns of U . Without loss or generality we may reorder the columns of A such that the rst k columns a1 , .

. . , ak of A form a basis for V A .

Now suppose that u 1 , . . .

, u k are linear dependent, that is, there exist constants c1 , . . .

, ck not all equal to zero such that kj=1 c j u j = 0. But then also kj=1 c j Bu j = kj=1 c j a j = 0, which by the linear independence of a1 , . .

. , ak implies that all the c j s are equal to zero..

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