ssrs barcode font free Because the null space contains a nonzero vector, there is no unique solution. in Software

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Because the null space contains a nonzero vector, there is no unique solution.
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{10, 1, 0} 3 3
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This is a particular solution.
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Linear Algebra
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The full set of solutions to the system is of the form t*nullspacebasis + particular where t is an arbitrary parameter. However, to express as a single list, we must first flatten nullspacebasis. generalsolution = t*Flatten[ nullspacebasis]+ particular
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{10 7 t, 1 + 5 t, 9 t} 3 3
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As a check, we substitute our general solution back into the original system. a.generalsolution // Expand {7, 2, 9}
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The Gauss-Jordan method for solving the linear system a.x = b is based upon the reduction of the augmented matrix [a|b] into reduced row echelon form by a series of elementary row operations. The three basic elementary row operations are: 1. interchanging two rows 2. multiplying a row by a non-zero constant 3. replacing one row by itself plus a multiple of another row It is easily seen that elementary row operations have no effect upon the solution of the system. A matrix is said to be in reduced row echelon form if 1. 2. 3. 4. all zero rows are placed at the bottom of the matrix each leading nonzero entry is 1 (called a leading 1) each entry above and below a leading 1 is 0 if two rows have leading 1s, the lower row has its leading 1 farther to the right
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To solve a linear system, we use elementary row operations to reduce the augmented matrix to reduced row echelon form. The solution(s) of the system, or the fact that no solution exists, may then be easily determined. Every student of linear algebra knows that row reduction is a time-consuming, tedious process that is highly prone to error. However, the Mathematica command RowReduce quickly reduces any matrix to reduced row echelon form.
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RowReduce[matrix] reduces matrix to reduced row echelon form.
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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
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