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12.23 Describe the set of vectors, S, spanned by (1, 2, 1, 2, 1), (1, 3, 2, 4, 2), and (1, 4, 3, 6, 3).
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a = {1, 2, 1, 2, 1}; b = {1, 3, 2, 4, 2}; c = {1, 4, 3, 6, 3}; m = {a, b, c}; m // MatrixForm 1 2 1 2 1 1 3 2 4 2 1 4 3 6 3 rref = RowReduce[m]; rref // MatrixForm 1 0 1 2 1 0 1 1 2 1 0 0 0 0 0 rref[[1]] {1, 0, 1, 2, 1} rref[[2]] {0, 1, 1, 2, 1} s * rref[[1]]+ t * rref[[2]] {s, t, s + t, 2 s + 2 t, s + t}
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Every vector in S is of this form.
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Form a matrix, m, using the given vectors as rows. The row space of m is the space spanned by a, b, and c. Then reduce the matrix to reduced row echelon form. The non-zero rows form a basis for the row space. Every vector in S is a linear combination of its basis vectors.
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12.24 A theorem of linear algebra says that every vector in the row space of A is orthogonal to every vector in the null space of A. Verify this result for
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1 2 3 4 6 7 8 9 A = 11 12 13 14 16 17 18 19 21 22 23 24
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5 10 15 20 25
It suffices to show that each basis vector of the row space is orthogonal to every basis vector in the null space. a = Partition[Range[25], 5]; a // MatrixForm 1 2 3 4 6 7 8 9 11 12 13 14 16 17 18 19 21 22 23 24 5 10 15 20 25
rowspacebasis = RowReduce[a]; rowspacebasis // MatrixForm 1 0 0 0 0 0 1 2 3 1 2 3 4 0 0 0 0 0 0 0 0 0 0 0 0
nullspacebasis = NullSpace[a]; nullspacebasis // MatrixForm 3 4 0 0 1 2 3 0 1 0 1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
To show that each vector in rowspacebasis is orthogonal to every vector in nullspacebasis, we must show that every dot product is 0. The easiest way is to multiply rowspacebasis by the transpose of nullspacebasis. The bottom three rows of zeros may be ignored.
rowspacebasis.Transpose[nullspacebasis] // MatrixForm
12.25 Construct a 5 5 matrix of random digits, a, and a 5 1 matrix of random digits, b, and solve the linear system ax = b using LinearSolve. Then verify that your solution is correct.
SOLUTION
a = Table[RandomInteger[9], {i, 1, 5}, {j, 1, 5}]; a // MatrixForm 0 7 0 5 3 9 7 6 8 0 5 5 6 4 8 0 9 2 3 1 2 3 6 9 5
Since the determinant 0, the system has a unique solution.
Det[a] 0 False
Linear Algebra
b = Table[RandomInteger[9], {i, 1, 5}]; b // MatrixForm 1 2 4 6 1 x = LinearSolve[a, b] 58 65 143 { 217, 651, 187, 111, 186} 651 434 a.x b True 1 2 3 A = 2 1 4 and b = 12.26 Solve the system Ax = b where 3 4 5
SOLUTION
14 12 , 13
9 10 17 , and 22 . 28 38
1 2 3 a = 2 1 4 ; 3 4 6
f = LinearSolve[a] LinearSolveFunction[{3,3}, <>]
f[{14, 12, 13}]
2, 3} 1, 3} 2, 4}
f[{9, 17, 28}]
f[{10, 22, 38}]
w + 2 x + 3 y + 3z = 12.27 Find the general solution of the system 2 w + x + 2 y + 5z = 2w + 2 x + y + 2z = SOLUTION 2w x 3 y + z =
1 2 3 2 1 2 a= 2 2 1 2 1 3 Det[a] 0 True nullspacebasis = NullSpace[a] {{ 13, 11, 10, 7}} particular = LinearSolve[a, b] 4 { 20, 7, 17, 0} 7 7 3 5 ; 2 1 b ={9,10, 7, 1};
9 10 7 1
Since the determinant is 0, we anticipate either no solution or an infinite number of solutions. Since the null space contains a non-zero vector, there will be an infinite number of solutions.
generalsolution = t * Flatten[nullspacebasis] + particular 4 { 20 13 t, 7 + 11 t, 17 10 t,7 t} 7 7