Problem 7.2. (a) Given two vectors in R2 of coordinates (c-a, c2-a2) and (b1-a b2-a2),prove that they are linearly dependent iff1 :|а1 b1 C1а2 b2 С2 — 0.11(ь) Given three vectors in R3 of coordinates (dj—a1,dg-a2, dҙ — аз), (с1—ај, С2— а9, Сз —аз),and (b a b - a2, b3 - a3), prove that theylinearly dependent iffareа1 bi C1a2 b2 C2 d2|аз бз Сз dз1d10.111

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Asked Oct 7, 2019
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Problem 7.2. (a) Given two vectors in R2 of coordinates (c-a, c2-a2) and (b1-a b2-a2),
prove that they are linearly dependent iff
1 :
|а1 b1 C1
а2 b2 С2 — 0.
1
1
(ь) Given three vectors in R3 of coordinates (dj—a1,dg-a2, dҙ — аз), (с1—ај, С2— а9, Сз —аз),
and (b a b - a2, b3 - a3), prove that they
linearly dependent iff
are
а1 bi C1
a2 b2 C2 d2
|аз бз Сз dз
1
d1
0.
1
1
1
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Problem 7.2. (a) Given two vectors in R2 of coordinates (c-a, c2-a2) and (b1-a b2-a2), prove that they are linearly dependent iff 1 : |а1 b1 C1 а2 b2 С2 — 0. 1 1 (ь) Given three vectors in R3 of coordinates (dj—a1,dg-a2, dҙ — аз), (с1—ај, С2— а9, Сз —аз), and (b a b - a2, b3 - a3), prove that they linearly dependent iff are а1 bi C1 a2 b2 C2 d2 |аз бз Сз dз 1 d1 0. 1 1 1

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Expert Answer

Step 1

A] Proof:

(с -а,с, - а,) and (b - a,b, - а,).
C1-a1
The vectors are
b-a
where A. B
Consider that the vectors are dependent, then A
b-a
c2-a2
are constants
The system of equations becomes
4(с, - а)+ в(b, -а) -о
4(с, - а,) +B(b, -а,) -0
The system is dependent as the vectors are linearly dependent, thus the determinant is
zero
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(с -а,с, - а,) and (b - a,b, - а,). C1-a1 The vectors are b-a where A. B Consider that the vectors are dependent, then A b-a c2-a2 are constants The system of equations becomes 4(с, - а)+ в(b, -а) -о 4(с, - а,) +B(b, -а,) -0 The system is dependent as the vectors are linearly dependent, thus the determinant is zero

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Step 2

Then,

G -а b-а
6, — а, b, -а,
(а -а)(6, - а,)-(ь, - a) (с, - а,) -0
сb, — са, - аb, +аа, -bс,+ba, + ас, — аа, -0
сb, - са, - аb, -bc, +ha, +ac, - о
0 =
Multiply by - sign and rearrange the terms
-сb, + Ga, +ab, +bc, - ba, - ac, -0
ab, - ас, - ba, +bc, + G,a, -c,b, %3D 0
a (b, - с,) -ь (а, +с,) +G (а, -b,) -0
b с
b C2
Clearly, the left hand side is the result of the determinant a,
1
1
1
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G -а b-а 6, — а, b, -а, (а -а)(6, - а,)-(ь, - a) (с, - а,) -0 сb, — са, - аb, +аа, -bс,+ba, + ас, — аа, -0 сb, - са, - аb, -bc, +ha, +ac, - о 0 = Multiply by - sign and rearrange the terms -сb, + Ga, +ab, +bc, - ba, - ac, -0 ab, - ас, - ba, +bc, + G,a, -c,b, %3D 0 a (b, - с,) -ь (а, +с,) +G (а, -b,) -0 b с b C2 Clearly, the left hand side is the result of the determinant a, 1 1 1

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Step 3
|-a (ь, -с,)-6 (а, -с,) +с (а, -b,).
That is, a, b, с,
1
1
1
a
b с
Thus a b c= 0.
1
1
1
|а, b с
Similarly, if a
c2= 0, then the vectors (c - a,c2-a,) and (-a,b-a2)
b
are
1
1
1
linearly dependent following some rearrangement in the computation.
Hence, it is proved that the vectors (c -ac2-a,) and (b-a.b,-a,) are linearly
b с
dependent iff a
b2c0
1
1
1
రో ఆ
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|-a (ь, -с,)-6 (а, -с,) +с (а, -b,). That is, a, b, с, 1 1 1 a b с Thus a b c= 0. 1 1 1 |а, b с Similarly, if a c2= 0, then the vectors (c - a,c2-a,) and (-a,b-a2) b are 1 1 1 linearly dependent following some rearrangement in the computation. Hence, it is proved that the vectors (c -ac2-a,) and (b-a.b,-a,) are linearly b с dependent iff a b2c0 1 1 1 రో ఆ

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