Questions and Answers
What must you know about the pivot columns in an augmented matrix to determine that the linear system is consistent and has a unique solution?
Every column in the augmented matrix except the rightmost column is a pivot column, and the rightmost column is not a pivot column.
Any list of five real numbers is a vector in R5.
True
The vector u results when a vector u - v is added to the vector v.
True
The weights c1,...,cp in a linear combination c1v1 + ... + cpvp cannot all be zero.
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When u and v are nonzero vectors, Span {u, v} contains the line through u and the origin.
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Asking whether the linear system corresponding to an augmented matrix [a1, a2, a3, b] has a solution amounts to asking whether b is in Span {a1, a2, a3}.
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What is the number of vectors in the set {a1, a2, a3}?
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Can each vector in R4 be written as a linear combination of the columns of the matrix A?
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Can a set of three vectors in R4 span all of R4?
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Why must the columns of matrix A span R3 if Ax = b has a unique solution?
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What fact allows you to conclude that the system Ax = w is consistent?
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What fact allows you to conclude that the system Ax = 4z is consistent?
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Explain why the solution to Ax = b is unique precisely when Ax = 0 has only the trivial solution.
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Find one nontrivial solution of Ax = 0 by inspection for A = [ 2 x 3 ].
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Explain why A(u + v) must be the zero vector given Au = 0 and Av = 0.
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Study Notes
Linear Systems and Augmented Matrices
- A linear system is consistent with a unique solution if every column in the augmented matrix (excluding the rightmost column) is a pivot column.
- In an augmented matrix, the rightmost column should not be a pivot column for consistency with a unique solution.
Vectors and Linear Combinations
- Any list of five real numbers qualifies as a vector in R⁵.
- Adding the vector u-v to vector v results in vector u.
- The weights in a linear combination can include all zeros, thus allowing for the possibility of a solution that is not constrained by non-zero weights.
- The span of nonzero vectors u and v includes the line connecting u to the origin.
Span and Solutions
- Asking if the linear system represented by an augmented matrix has a solution relates to whether the vector b is included in the Span of vectors a₁, a₂, and a₃.
- A set of three vectors in R⁴ cannot span all of R⁴ since a matrix made up of these vectors lacks enough pivot positions for every dimension.
- A unique solution for Ax = b indicates that all columns of A are pivot columns, thus spanning R³.
Theorem Applications in Consistency
- The linear combination of two vector solutions results in a valid solution for the system Ax = w, confirming consistency.
- For the system Ax = 4z to be consistent, it is derived from a similar previous solution, applying scalar multiplication.
Uniqueness of Solutions
- A solution to Ax = b is unique if Ax = 0 only has the trivial solution, which occurs when every column is a pivot column.
- Nontrivial solutions to Ax = 0 can be identified through relationships between the matrix's columns and independent choices for vector components.
Properties of Linear Transformations
- If vectors u and v satisfy Au = 0 and Av = 0, their sum A(u + v) remains a zero vector, leading to the conclusion that any linear combination A(cu + dv) also results in zero.
These notes should provide clear insights into concepts surrounding linear systems, augmented matrices, and vector relationships, as well as implications regarding solution uniqueness and consistency.
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Test your knowledge on key concepts in linear algebra through these flashcards. Each card includes essential definitions and true/false questions related to vectors and augmented matrices. Challenge yourself to understand the foundations of linear system solutions!