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    Linear algebra test true and false

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    a homogeneous equation is always consistent
    TRUE Ax=b always has “trivial” solution (where all the variables are 0)
    the homogeneous equation Ax=0 has the trivial solution if and only if the equation has at least one free variable
    FALSE Ax=0 always has the trivial solution
    the equation Ax=0 gives an explicit description of its solution set
    FALSE the original equation is the implicit description and the explicit description is the equation solved for the span
    the equation x=p + tv describes a line through v parallel to p
    FALSE describes a line through p parallel to v
    a homogeneous system of equations can be inconsistent
    FALSE always has to have one solution x=0
    if x is a nontrivial solution of Ax=0, then every entry in x is nonzero
    FALSE, can have some 0 entries where x=0 but not all
    the equation Ax=b is a homogeneous if the zero vector is a solution
    TRUE the zero vector is always a solution to homogeneous systems
    the columns of A are linearly independent if the equation Ax=0 has the trivial solution
    FALSE a homogeneous system always has the trivial solution
    if S is a linearly dependent set, then each vector is a linear combination of the other vectors S
    FALSE not every vector in a linearly dependent set is a combination of the preceding vectors
    the columns of any 4×5 matrix are linearly dependent
    TRUE is there are more columns than rows then its linearly dependent
    if x and y are linearly independent, and if {x, y, z} is linearly dependent, then z is in Span {x,y}
    TRUE {x, y, z} will be linearly dependent if and only if w is on the plane spanned by u and v
    if u and v are linearly independent, and if w is in the span {u, v}, then {u, v, w} is linearly dependent
    TRUE S={v1…vp} of 2 or more vectors is linearly dependent if and only if at least one of the vectors in S is a linear combination of the others
    if three vectors in R3 lie in the same plane in R3, then they are linearly dependent
    TRUE the third vector must be a multiple of the first or the second vector (who can’t be multiples of each other which means they are linearly independent)
    if a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent
    FALSE set consisting of

    v1=[1, -2, 3] v2= [2, -4, 6]

    is linearly dependent, the theorem p>n does not apply when p

    if a set in Rn is linearly dependent, then the set contains more than n vectors
    FALSE can have the same number of vectors as n
    v1= [3, 1] v2=[6, 2]
    v2 is a multiple of v1 so {v1, v2} is linearly dependent and has the same number of vectors
    if v1…v4 are in R4 and v3=2v1+v2 then {v1, v2, v3, v4} is linearly dependent
    TRUE S={v1…vp} of 2 or more vectors is linearly dependent if and only if at least one of the vectors in S is a linear combination of the others
    if v1 and v2 are in R4 and v2 is not a scalar multiple of v1, then {v1,v2} is linearly dependent
    FALSE the vector v1 could be the zero vector
    if v1,…v5 are in R5 and v3=0 then {v1, v2, v3, v4, v5} is linearly dependent
    TRUE if a set S={v1…vp} in Rn contains the zero vector, then the set is linearly dependent
    if v1, v2, v3 are in R3 and v3 is not a linear combination of v1, v2 then {v1, v2, v3} is linearly independent
    FALSE v1 and v2 can be multiples of each other making the system linearly dependent
    if v1…v4 is in R4 and {v1,v2,v3} is linearly dependent, then {v1, v2, v3, v4} is also linearly dependent
    TRUE a linear dependence relation among v1, v2, v3 may be extended to linear dependence relation among v1, v2, v3, v4 by placing a zero weight on v4
    if {v1…v4} is a linearly independent set of vectors in R4 then {v1, v2, v3} is also linearly independent [think about x1v1+x2v2+x3v3+0*v4=0]
    TRUE if the equation x1v1+x2v2+x3v3+0 x v4=0 had a normal solution with at least one of the other three vectors being nonzero, then so would the equation x1v1+x2v2+x3v3+0 x v4=0. but that cannot happen because {v1, v2, v3, v4} is linearly independent. so {v1, v2, v3} must be linearly independent.
    a linear transformation is a special type of function
    TRUE a linear transformation is a function with certain properties
    if A is a 3 x 5 matrix and T is a transformation defined by T(x)=Ax then the domain of T is R3
    FALSE the domain is R5, the domain of T is the number of columns
    if A is an m x n matrix, then the range of the transformation x |–> Ax is Rm
    FALSE the range is the set of all linear combinations of the columns of A, because each image T(x) is of the form Ax
    every linear transformation is a matrix transformation
    FALSE every matrix transformation is a linear transformation, but not the reverse
    a transformation T is linear if and only if T(c1v1+c2v2)=c1T(v1)+c2T(v2) for all v1 and v2 in the domain of T and for all scalars c1 and c2
    TRUE linear transformations preserve the operations of vector addition and scalar multiplication
    the range of the transformation x |–> Ax is the set of all linear combinations of the columns of A
    TRUE the range is the set of all linear combinations of the columns of A, because each image T(x) is of the form Ax
    every matrix transformation is a linear transformation
    TRUE every matrix transformation is a linear transformation, but not the reverse
    if T: Rn –> Rm is a linear transformation and if c is in Rm, then a uniqueness question is “Is c in the range of T?”
    FALSE this is an existence question, another way of asking if Ax=C is consistent
    A linear transformation preserves the operations of vector addition and scalar multiplication
    TRUE linear transformations preserve the operations of vector addition and scalar multiplication
    a linear transformation T: Rn –> Rm always maps the origin of Rn to the origin of Rm
    TRUE If T is a linear transformation then T(0)=0, and T(cu+dv)=cT(u)+dT(v)

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    Linear algebra test true and false. (2018, Oct 20). Retrieved from https://artscolumbia.org/linear-algebra-test-true-and-false-3-37361-61293/

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