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Linear Algebra Controls

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Submitted By tochitemi
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There is ONE degree of freedom and the system is consistent. b. The system is NOT consistent. c. The system is consistent and there is ONE degree of freedom. d. The system is consistent and there is ONE degree of freedom. e. The system is consistent but there is NO degree of freedom. f. The system is consistent and there are TWO degree of freedom.
2. Ab = 1 1 0 -1 2 1 1 1 1 2 0 2 -1 -1 2 5 -5 0 There is ONE free parameter [in 2nd column] 2 2 1 -2 5 0 ans = 1 1 0 0 0 3 0 0 1 0 0 -1 0 0 0 1 0 0 0 0 0 0 1 -1
3 a. 1. A=[1 -2 3 -6; 3 0 -2 5; 1 1 -2 5]; rref(A) 1 0 0 1 0 1 0 2 0 0 1 -1
Yes there is a Linear combination: p= q1+2q2-q3 2. A=[1 -2 3 1; 3 0 -2 0; 1 1 -2 0]; rref(A) 1.0000 0 0 0.6667 0 1.0000 0 1.3333 0 0 1.0000 1.0000
Yes there is a Linear Combination: p= 0.67q1 + 1.333q2 + q3
3b
1. A=[1 2 0 2; 2 0 4 1; 0 3 -3 1; 1 -1 3 1]; rref(A)

1 0 2 0 0 1 -1 0 0 0 0 1 0 0 0 0 The given p row is not a linear combination of the q vectors as they do not span all elements in p.

2. A=[1 2 0 1; 2 0 4 18; 0 3 -3 -12; 1 -1 3 13]; rref(A) 1 0 2 9 0 1 -1 -4 0 0 0 0 0 0 0 0 The given p row is a linear combination of the q vectors as they = span all elements in p

3c 1. A=[2 1 1 1; 1 -1 2 0; -3 0 0 0; 4 1 0 0]; rref(A) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
P is not linear combination of Q 2. A=[2 1 1 0; 1 -1 2 1; -3 0 0 0; 4 1 0 0]; rref(A) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
P is not linear combination of Q

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