Math 308 Midterm T/F

8 September 2022
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question
A linear system with three equations and two variables must be inconsistent.
answer
False
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A linear system with three equations and five variables must be consistent.
answer
False.
question
A system in echelon form can have more variables than equations.
answer
True.
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If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.
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True, by definition of equivalent matrices.
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Different sequences of row operations can lead to different echelon forms for the same matrix.
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True. For example, 2 4 āˆ’(1/2)R1 + R2 => R2~ 1 1 2 4 0 āˆ’1 And 2 4 1 1 R1 ā‡” R2~ 1 1 2 4 āˆ’2R1 + R2 right double arrow implies R2~ 1 1 0 2
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Different sequences of row operations can lead to different reduced echelon forms for the same matrix.
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False, by Theorem 1.5.
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If a linear system has four equations and seven variables, then it must have infinitely many solutions.
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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
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Every linear system with free variables has infinitely many solutions.
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True, a free variable can take any value, and so there are infinitely many solutions.
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Any linear system with more variables than equations cannot have a unique solution.
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True. If it is consistent, there will be at least one free variable, and hence infinitely many solutions.
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If a linear system has the same number of equations and variables, then it must have a unique solution.
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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.
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A vector can have positive or negative components, but a scalar must be positive.
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False. Scalars may be any real number, such as c = āˆ’1.
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If c1 and c2 are scalars and u is a vector, then (c1 + u)c2 = c1c2 + c2u.
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False. The sum c1 + u of a scalar and a vector is undefined.
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The vectors [1,āˆ’4, 5] and [āˆ’2, 8,10] point in opposite directions.
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False. They do not point in opposite directions, as there does not exist c < 0 such that [1, āˆ’4, 5] = c * [āˆ’2,8, 10]
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The vector 2u is longer than the vector āˆ’6u.
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False. For example, the length of 2*[1, 0] = [2, 0] is 2, but the length of (āˆ’6)*[1, 0] = [āˆ’6, 0] is 6.
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If a set of vectors includes 0, then it cannot span Rn.
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False, the zero vector can be included with any set of vectors which already span Rn.
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Suppose A is a matrix with n rows and m columns. If n < m, then the columns of A span Rn.
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False, since every column of A may be a zero column.
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Suppose A is a matrix with n rows and m columns. If m < n, then the columns of A span Rn.
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False
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If A is a matrix with columns that span Rn, then Ax = b has a solution for all b in Rn.
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True
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If {u1, u2, u3} spans R3, then so does {u1, u2, u3, u4}.
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True, the span of a set of vectors can only increase (with respect to set containment) when adding a vector to the set.
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If {u1, u2, u3} does not span R3, then neither does {u1, u2, u3, u4}.
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False. Consider u1 = (0, 0, 0), u2 = (1, 0, 0), u3 = (0, 1, 0), and u4 = (0, 0, 1).
question
If {u1, u2, u3, u4} does not span R3, then neither does {u1, u2, u3}.
answer
True. The span of {u1, u2, u3} will be a subset of the span of {u1, u2, u3, u4}.
question
If u4 is a linear combination of {u1, u2, u3}, then span{u1, u2, u3, u4} = span{u1, u2, u3}.
answer
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}.
question
If u4 is not a linear combination of {u1, u2, u3}, then span{u1, u2, u3, u4} ā‰  span{u1, u2, u3}.
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True, since u4 is in span{u1, u2, u3, u4}, but u4 not in span{u1, u2, u3}.
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Linearly dependent.
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If {u1, u2, , um} is a set of vectors in Rn and n < m, then the set is linearly dependent.
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Linearly independent.
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The vectors are not scalar multiples of each other.
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Linearly dependent.
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Any collection of vectors containing the zero vector must be linearly dependent.
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If a set of vectors in Rn is linearly dependent, then the set must span Rn.
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False. For example, u = (1, 0) and v = (2, 0) are linearly dependent but do not span R2.
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Linearly dependent.
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If m > n, then a set of m vectors in Rn is linearly dependent.
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If A is a matrix with more columns than rows, then the columns of A are linearly independent.
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False. For example, A = [1 2 3] [0 0 0] has more columns than rows, but the columns are linearly dependent.
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If A is a matrix with linearly independent columns, then Ax = b has a solution for all b.
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False. For example, if A = [1, 1] and b = [1, 0], then Ax = b has no solution.
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If {u1, u2, u3} is linearly dependent, then so is {u1, u2, u3, u4}.
answer
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.
question
If {u1, u2, u3, u4} is linearly independent, then so is {u1, u2, u3}.
answer
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.
question
If {u1, u2, u3, u4} is linearly dependent, then so is {u1, u2, u3}.
answer
False. Consider u1 = (1, 0, 0), u2 = (0, 1, 0), u3 = (0, 0, 1), u4 = (0, 0, 0).
question
If u4 is a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly independent.
answer
False. If u4 = x1u1 + x2u2 + x3u3, then x1u1 + x2u2 + x3u3 āˆ’ u4 = 0, and since the coefficient of u4 is āˆ’1, {u1, u2, u3, u4} is linearly dependent.
question
If u4 is not a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly independent.
answer
False. Consider u1 = (1, 0, 0), u2 = (1, 0, 0), u3 = (1, 0, 0), u4 = (0, 1, 0).
question
If u4 is not a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly dependent.
answer
False. Consider u1 = (1, 0, 0, 0), u2 = (0, 1, 0, 0), u3 = (0, 0, 1, 0), u4 = (0, 0, 0, 1).