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ssrs barcode font free Because the null space contains a nonzero vector, there is no unique solution. in Software
Because the null space contains a nonzero vector, there is no unique solution. QR Code Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Create QR Code In None Using Barcode drawer for Software Control to generate, create QRCode image in Software applications. {10, 1, 0} 3 3 Decode QRCode In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Print QR Code In C#.NET Using Barcode drawer for .NET framework Control to generate, create QR Code image in .NET applications. This is a particular solution.
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EAN / UCC  13 Creation In ObjectiveC Using Barcode creation for iPhone Control to generate, create EAN128 image in iPhone applications. Matrix Barcode Drawer In .NET Framework Using Barcode creation for VS .NET Control to generate, create Matrix 2D Barcode image in VS .NET applications. EXAMPLE 17 Determine the reduced row echelon form of the 4 5 matrix whose general entry ai,j =  i j . a = Table[Abs[i j], {i, 1, 4}, {j, 1, 5}]; a // MatrixForm 0 1 2 3 1 0 1 2 2 1 0 1 3 2 1 0 4 3 2 1 RowReduce[a] 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 1 We now illustrate how row reduction can be used to solve a linear system. For comparison purposes we use the three examples previously considered in Examples 15 and 16. Linear Algebra
EXAMPLE 18
(a) 2 x + y + z = 7, x 4 y + 3 z = 2, 3 x + 2 y + 2 z = 13 (b) 2 x + y + z = 7, x 4 y + 3 z = 2, 3 x 3 y + 4 z = 13 (c) 2 x + y + z = 7, x 4 y + 3 z = 2, 3 x 3 y + 4 z = 9 (unique solution) (no solution) (infinite number of solutions) We find the augmented matrix for each of the three systems. 2 1 1 7 a1 = 1 4 3 2 ; 3 2 2 13 2 1 1 7 a2 = 1 4 3 2 ; 3 3 4 13 2 1 1 7 a3 = 1 4 3 2 ; 3 3 4 9 RowReduce[a1] //MatrixForm 1 0 0 1 0 1 0 2 0 0 1 3 RowReduce[a2] //MatrixForm
7 1 0 9 0 0 1 5 0 9 0 0 0 1 This reduced matrix, when interpreted as a system of equations, reads: x = 1, y = 2, z = 3.
The bottom row reads 0x + 0y + 0z = 1, which, of course, is impossible. This contradiction (a row of 0s and a nal 1) reveals that no solution is possible. RowReduce[a3] //MatrixForm
7 1 0 9 0 1 5 9 0 0 0 10 3 1 3 0 The bottom row of 0s is not a contradiction. However, there cannot be a unique solution. If we let z = t, an independent parameter, the solution may be put into the form

