Search
Create
Log in
Sign up
Log in
Sign up
Get ahead with a $300 test prep scholarship
| Enter to win by Tuesday 9/24
Learn more
Math 308 Midterm T/F
STUDY
Flashcards
Learn
Write
Spell
Test
PLAY
Match
Gravity
Terms in this set (36)
A linear system with three equations and two variables must be inconsistent.
False
A linear system with three equations and five variables must be consistent.
False.
A system in echelon form can have more variables than equations.
True.
If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.
True, by definition of equivalent matrices.
Different sequences of row operations can lead to different echelon forms for the same matrix.
True. For example,
2 4 −(1/2)R1 + R2 => R2~
1 1
2 4
0 −1
And
2 4
1 1 R1 ⇔ R2~
1 1
2 4 −2R1 + R2 right double arrow implies R2~
1 1
0 2
Different sequences of row operations can lead to different reduced echelon forms for the same matrix.
False, by Theorem 1.5.
If a linear system has four equations and seven variables, then it must have infinitely many solutions.
False,
It could be inconsistent, and therefore have no solutions, as with the system
x1+x2+x3+x4 +x5+x6+x7=0
x1+x2+x3 =1
x4 +x5 =1
x6+ x7=1
Every linear system with free variables has infinitely many solutions.
True, a free variable can take any value, and so there are infinitely many solutions.
Any linear system with more variables than equations cannot have a unique solution.
True. If it is consistent, there will be at least one free variable, and hence infinitely many solutions.
If a linear system has the same number of equations and variables, then it must have a unique solution.
False. For example, the system
x1 + x2 = 0
x1 + x2 = 1
has no solutions. And the system
x1 + x2 = 1
2x1 + 2x2 = 2
has infinitely many solutions.
A vector can have positive or negative components, but a scalar must be positive.
False. Scalars may be any real number, such as
c = −1.
If c1 and c2 are scalars and u is a vector, then
(c1 + u)c2 = c1c2 + c2u.
False. The sum c1 + u of a scalar and a vector is undefined.
The vectors [1,−4, 5] and [−2, 8,10] point in opposite directions.
False. They do not point in opposite directions, as there does not exist c < 0 such that [1, −4, 5] = c * [−2,8, 10]
The vector 2u is longer than the vector −6u.
False. For example, the length of 2
[1, 0] = [2, 0] is 2, but the length of (−6)
)*[1, 0] = [−6, 0] is 6.
If a set of vectors includes 0, then it cannot span
Rn.
False, the zero vector can be included with any set of vectors which already span Rn.
Suppose A is a matrix with n rows and m columns. If
n < m, then the columns of A span Rn.
False, since every column of A may be a zero column.
Suppose A is a matrix with n rows and m columns. If
m < n, then the columns of A span Rn.
False
If A is a matrix with columns that span Rn, then
Ax = b has a solution for all b in Rn.
True
If {u1, u2, u3} spans R3, then so does {u1, u2, u3, u4}.
True, the span of a set of vectors can only increase (with respect to set containment) when adding a vector to the set.
If {u1, u2, u3} does not span R3, then neither does {u1, u2, u3, u4}.
False. Consider u1 = (0, 0, 0), u2 = (1, 0, 0), u3 = (0, 1, 0), and u4 = (0, 0, 1).
If {u1, u2, u3, u4} does not span R3, then neither does {u1, u2, u3}.
True. The span of {u1, u2, u3} will be a subset of the span of {u1, u2, u3, u4}.
If u4 is a linear combination of {u1, u2, u3}, then
span{u1, u2, u3, u4} = span{u1, u2, u3}.
True. Since u4 is a linear combination of {u1, u2, u3}, any vector in span{u1, u2, u3, u4} can be written as a linear combination of {u1, u2, u3}.
If u4 is not a linear combination of {u1, u2, u3}, then
span{u1, u2, u3, u4} ≠ span{u1, u2, u3}.
True, since u4 is in span{u1, u2, u3, u4}, but u4 not in span{u1, u2, u3}.
Linearly dependent.
If {u1, u2, , um} is a set of vectors in Rn and n < m, then the set is linearly dependent.
Linearly independent.
The vectors are not scalar multiples of each other.
Linearly dependent.
Any collection of vectors containing the zero vector must be linearly dependent.
If a set of vectors in Rn is linearly dependent, then the set must span Rn.
False. For example, u = (1, 0) and v = (2, 0) are linearly dependent but do not span R2.
Linearly dependent.
If m > n, then a set of m vectors in Rn is linearly dependent.
If A is a matrix with more columns than rows, then the columns of A are linearly independent.
False. For example, A = [1 2 3]
[0 0 0]
has more columns than rows, but the columns are linearly dependent.
If A is a matrix with linearly independent columns, then
Ax = b has a solution for all b.
False. For example, if A = [1, 1] and b = [1, 0], then Ax = b has no solution.
If {u1, u2, u3} is linearly dependent, then so is {u1, u2, u3, u4}.
True. If {u1, u2, u3} is linearly dependent, then the equation x1u1 + x2u2 + x3u3 = 0 has a nontrivial solution, and therefore so does x1u1 + x2u2 + x3u3 + x4u4 = 0.
If {u1, u2, u3, u4} is linearly independent, then so is
{u1, u2, u3}.
True. If {u1, u2, u3, u4} is linearly independent, then the equation x1u1 + x2u2 + x3u3 + x4u4 = 0 has only the trivial solution, and therefore so does x1u1 + x2u2 + x3u3 = 0.
If {u1, u2, u3, u4} is linearly dependent, then so is
{u1, u2, u3}.
False. Consider u1 = (1, 0, 0), u2 = (0, 1, 0), u3 = (0, 0, 1), u4 = (0, 0, 0).
If u4 is a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly independent.
False. If u4 = x1u1 + x2u2 + x3u3, then x1u1 + x2u2 + x3u3 − u4 = 0, and since the coefficient of u4 is −1, {u1, u2, u3, u4} is linearly dependent.
If u4 is not a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly independent.
False. Consider u1 = (1, 0, 0), u2 = (1, 0, 0), u3 = (1, 0, 0), u4 = (0, 1, 0).
If u4 is not a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly dependent.
False. Consider u1 = (1, 0, 0, 0), u2 = (0, 1, 0, 0), u3 = (0, 0, 1, 0), u4 = (0, 0, 0, 1).
;