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Linear Algebra
Linear Algebra: True or False Questions (Part 1)
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Terms in this set (26)
Every matrix is row-equivalent to a unique matrix in echelon form.
False
This is true if we replace "echelon form" with "reduced echelon form."
Any system of n linear equations in n variables has at most n solutions.
False
If the system has a free variable then it has infinitely many solutions.
If a system of linear equations has two different solutions, it must have infinitely many solutions.
True
If Ax_1 = b and Ax_2 = b, then A(x_1 - x_2) = b-b = 0, so any multiple of x_1 + x_2 can be added to any solution of Ax = b to get another solution.
If a system of linear equations has no free variables then it has a unique solution.
False
This would be true if the system were consistent, but it may not have a solution at all.
If an augmented matrix [A | b] is transformed into [C | d] by elementary row operations, then the equations Ax = b and Cx = d have exactly the same solution sets.
True
Elementary operations preserve solution sets.
If a system Ax = b has more than one solution, then so does the system Ax = 0.
True
Suppose Ax_1 = b and Ax_2 = b for two different vectors x_1 and x_2. Then A(x_1 - x_2) = Ax_1 - Ax_2 = b - b = 0, and likewise A(x_2 - x_1) = 0, but x_1 - x_2 does not equal x_2 - x_1.
If A is an m x n matrix and the equation Ax = b is consistent for some b, then the columns of A span R^m.
False
The columns of A span R^m if and only if the equation Ax = b is consistent for all vectors b in R^m.
If an augmented matrix [A | b] can be transformed by elementary row operations in reduced echelon form, then the equation Ax = b is consistent.
False
The system is consistent if and only if the reduced echelon form of the augmented matrix does not contain a row of the form [0 0 ... 0 b] with b not equal to zero.
If matrices A and B are row-equivalent, they have the same reduced echelon form.
True
Since A and B are row-equivalent, A can be row-reduced to obtain B, which can then be row-reduced to its reduced echelon form which is unique.
The equation Ax = 0 has the trivial solution if and only if there are no free variables.
False
The equation Ax = 0 always has the trivial solution (which means specifically x = 0) since A0 = 0 for any matrix A.
If A is an m x n matrix and the equation Ax = b is consistent for every b in R^m, then A has m pivot columns.
True
A has a pivot position in every row, and since A has m rows this means that A has m pivots and thus m pivot columns.
If an m x n matrix A has a pivot position in every row, then the equation Ax = b has a unique solution for each b in R^m.
False
Ax = b has a solution for each b in R^m, but this solution may not be unique (and in fact will not be if the system has a free variable).
If an n x n matrix A has n pivot positions, then the reduced echelon form of A is the n x n identity matrix.
True
Since there can be at most one pivot in each row and one pivot in each column and there are n pivots, they must be on the main diagonal. Since the pivots in the reduced echelon form must be ones, the reduced echelon form must be the identity matrix.
If A and B are 3 x 3 matrices and each has three pivot positions, then A can be transformed into B by elementary row operations.
True
With three pivot positions (one in each row and one in each column), the reduced echelon form of A and B must be the 3 x 3 identity matrix I. Since A and B are row-equivalent to I, they are row-equivalent to each other.
If A is an m x n matrix, if the equation Ax = b has at least two different solutions, and if the equation Ax = c is consistent, then the equation Ax = c has (infinitely) many solutions.
True
Since Ax = b has at least two solutions, there must be a free variable in A.
If A and B are row-equivalent m x n matrices and if the columns of A span R^m, then so do the columns of B.
True
A has a pivot position in every row; since B is row-equivalent to A it has the same reduced echelon form, and therefore also has a pivot position in every row. Therefore the columns of B also span R^m.
If none of the vectors in the set S = {v_1, v_2, v_3} in R^3 is a multiple of one of the other vectors, then S is linearly independent.
False
The set S = {e_1, e_2, e_1 + e_2} gives a counterexample.
If {u, v, w} is linearly independent, then u, v, and w are not in R^2.
True
If u, v, and w were vectors in R^2, the set {u, v, w} would consist of more vectors (3) than the number of entries in each vector (2), which would be linearly dependent.
In some cases it is possible for four vectors to span R^5.
False
If this were true, then the matrix A having these vectors as its columns would be 5 x 4. But then its reduced echelon form could have at most 4 pivots (one in each column), and therefore could not have a pivot in each row, so the columns of A could not span R^5.
If u and v are in R^m, then -u is in Span {u, v}.
True
Since -u = (-1)
u + (0)
v it is a linear combination of u and v, and thus is in Span {u, v}, which is the set of all such linear combinations.
If u, v, and w are nonzero vectors in R^2, then w is a linear combination of u and v.
False
If u and v are vectors on the same line (for example, if v = 2*u) and w is not on that same line, then w is not a linear combination of u and v.
If w is a linear combination of u and v in R^n, then u is a linear combination of v and w.
False
If w = 0
u + 1
v = v and u is not a multiple of v, then u cannot be a linear combination of v and w.
Suppose that v_1, v_2, and v_3 are in R^5, v_2 is not a multiple of v_1, and v_3 is not a linear combination of v_1 and v_2. Then {v_1, v_2, v_3} is linearly independent.
False
If v_1 = 0, and v_2 and v_3 do not, the set {v_1, v_2, v_3} is linearly dependent since it contains the zero vector.
A linear transformation is a function.
True
It is a function which has vectors as input and output.
If A is a 6 x 5 matrix, the linear transformation x --> Ax cannot map R^5 onto R^6.
True
The transformation is onto R^6 if and only if the columns of A span R^6. But A has only five columns so it can have at most five pivot positions, and thus cannot have a pivot position in each row. Therefore the columns of A cannot span R^6.
If A is an m x n matrix with m pivot columns, the linear transformation x --> Ax is a one-to-one mapping.
False
The transformation is one-to-one if and only if the columns of A are linearly independent. However, if n > m they cannot be linearly independent.
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