For #3-8: True or False. If false, explain why. 3. If V1, V2, V3, V4 € IR4 and v3 = 2v1 + V2, then {V1, V2, V3, V4} is linearly dependent. this is true, Vs is a linear combination of 1₁ and 1₂ 4. If V₁, V2, V3, V4 € R4 and v₂ = 0, then {V₁, V2, V3, V4} is linearly independent. This is false bes a zero vector with make 1 column unable to have a which will have a free variable =) Linearly Dependent 5. If v₁ and v₂ are nonzero vectors in R5 and v₂ is not a scalar multiple of V₁, then {V₁, V₂ } is linearly independent. 6. If V₁, V2, V3, V4 € R4 and v₂ is not a linear combination of V₁, V3, V4, then {V₁, V2, V3, V4} is linearly independent. This is false, V3 V4, on v₁ can all be linear combinations of each other pivot 7. If V1, V2, V3, V4 € R4 and {V₁, V2, V3} is linearly dependent, then {V₁, V2, V3, V4} is also linearly dependent. 8. If V₁, V2, V3, V4 € R4 and {V₁, V2, V3, V4} is linearly independent, then {V₁, V2, V3 } is also linearly independent. This is false bes it is possible V4 oan be a linear combination of the other vectors (V₁, V2103) 9. Let A be an mxn matrix, and let u,ve Rn, such that Au = 0 and Av = 0. Do not skip steps. State theorems used in supporting reasoning. (b) Show that A(cu + dv) = 0 for scalars c and d. Let Autò and A=0 A'((ut + d) where C, and I are scalars = A. (cu) + A (du) (Matrix multiplication distribute over addition ²)+(AF) d (a) Show that A(u + v) must be the zero vector.
For #3-8: True or False. If false, explain why. 3. If V1, V2, V3, V4 € IR4 and v3 = 2v1 + V2, then {V1, V2, V3, V4} is linearly dependent. this is true, Vs is a linear combination of 1₁ and 1₂ 4. If V₁, V2, V3, V4 € R4 and v₂ = 0, then {V₁, V2, V3, V4} is linearly independent. This is false bes a zero vector with make 1 column unable to have a which will have a free variable =) Linearly Dependent 5. If v₁ and v₂ are nonzero vectors in R5 and v₂ is not a scalar multiple of V₁, then {V₁, V₂ } is linearly independent. 6. If V₁, V2, V3, V4 € R4 and v₂ is not a linear combination of V₁, V3, V4, then {V₁, V2, V3, V4} is linearly independent. This is false, V3 V4, on v₁ can all be linear combinations of each other pivot 7. If V1, V2, V3, V4 € R4 and {V₁, V2, V3} is linearly dependent, then {V₁, V2, V3, V4} is also linearly dependent. 8. If V₁, V2, V3, V4 € R4 and {V₁, V2, V3, V4} is linearly independent, then {V₁, V2, V3 } is also linearly independent. This is false bes it is possible V4 oan be a linear combination of the other vectors (V₁, V2103) 9. Let A be an mxn matrix, and let u,ve Rn, such that Au = 0 and Av = 0. Do not skip steps. State theorems used in supporting reasoning. (b) Show that A(cu + dv) = 0 for scalars c and d. Let Autò and A=0 A'((ut + d) where C, and I are scalars = A. (cu) + A (du) (Matrix multiplication distribute over addition ²)+(AF) d (a) Show that A(u + v) must be the zero vector.
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter1: Equations And Graphs
Section1.3: Lines
Problem 92E
Related questions
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I need.help with question 5 and 7 please
![For #3-8: True or False. If false, explain why.
3. If V₁, V2, V3, V4 € R4 and v3 = 2V₁ + V2, then {V₁, V2, V3, V4} is linearly dependent.
this is true, V₂ is a linear combination of 1₁ and 1₂
4. If V₁, V2, V3, V4 € R4 and v₂ = 0, then {V₁, V2, V3, V4} is linearly independent.
This is false bes a zero vectar with make 1 column unable to have a pivot
which will have a free variable => Linearly Dependent
5. If v₁ and v₂ are nonzero vectors in IRS and v₂ is not a scalar multiple of v₁, then {V₁, V₂ } is linearly
independent.
6. If V₁, V2, V3, V4 € R4 and v₂ is not a linear combination of V1, V3, V4, then {V₁, V2, V3, V4} is linearly independent.
This is false, V3, V4, on v₁ can all be linear combinations of each other
7. If V₁, V2, V3, V4 € R4 and {V₁, V2, V3 } is linearly dependent, then {V₁, V2, V3, V4} is also linearly dependent.
8. If V₁, V2, V3, V4 € R4 and {V₁, V2, V3, V4} is linearly independent, then {V1, V2, V3 } is also linearly independent.
This is false bes it is possible V4 oan be a
be a linear combination of the other vectors
(V₁, V21 U₂)
9. Let A be an mxn matrix, and let u,ve Rn, such that Au = 0 and Av = 0.
(a) Show that A(u + v) must be the zero vector.
Do not skip steps. State theorems used in supporting reasoning.
(b) Show that A (cu + dv) = 0 for scalars c and d.
Let Au= o and Av = 0
A' (Cu + du) where C, and I are scalars
= A· (cu`` ) + A'(du) (Matrix maltiplication
(distribute over additiony
=((Au) +d.(AU)
= c⋅0+d⋅0
0
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65ce4946-20f3-402b-8a73-35e62e74d93e%2Fb78cbe81-4093-42f3-b204-52b2282d272b%2Ftopamk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For #3-8: True or False. If false, explain why.
3. If V₁, V2, V3, V4 € R4 and v3 = 2V₁ + V2, then {V₁, V2, V3, V4} is linearly dependent.
this is true, V₂ is a linear combination of 1₁ and 1₂
4. If V₁, V2, V3, V4 € R4 and v₂ = 0, then {V₁, V2, V3, V4} is linearly independent.
This is false bes a zero vectar with make 1 column unable to have a pivot
which will have a free variable => Linearly Dependent
5. If v₁ and v₂ are nonzero vectors in IRS and v₂ is not a scalar multiple of v₁, then {V₁, V₂ } is linearly
independent.
6. If V₁, V2, V3, V4 € R4 and v₂ is not a linear combination of V1, V3, V4, then {V₁, V2, V3, V4} is linearly independent.
This is false, V3, V4, on v₁ can all be linear combinations of each other
7. If V₁, V2, V3, V4 € R4 and {V₁, V2, V3 } is linearly dependent, then {V₁, V2, V3, V4} is also linearly dependent.
8. If V₁, V2, V3, V4 € R4 and {V₁, V2, V3, V4} is linearly independent, then {V1, V2, V3 } is also linearly independent.
This is false bes it is possible V4 oan be a
be a linear combination of the other vectors
(V₁, V21 U₂)
9. Let A be an mxn matrix, and let u,ve Rn, such that Au = 0 and Av = 0.
(a) Show that A(u + v) must be the zero vector.
Do not skip steps. State theorems used in supporting reasoning.
(b) Show that A (cu + dv) = 0 for scalars c and d.
Let Au= o and Av = 0
A' (Cu + du) where C, and I are scalars
= A· (cu`` ) + A'(du) (Matrix maltiplication
(distribute over additiony
=((Au) +d.(AU)
= c⋅0+d⋅0
0
4
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