Suppose Ax b has a solution. Explain why the solution is unique precisely when Ax = 0 has only the trivial solution.
Suppose Ax b has a solution. Explain why the solution is unique precisely when Ax = 0 has only the trivial solution.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:**Question:**
Suppose \( Ax = b \) has a solution. Explain why the solution is unique precisely when \( Ax = 0 \) has only the trivial solution.
**Explanation:**
In linear algebra, the equation \( Ax = b \) represents a system of linear equations where \( A \) is the coefficient matrix, \( x \) is the column vector of variables, and \( b \) is the column vector of constants.
A solution to \( Ax = b \) is unique if and only if the homogeneous equation \( Ax = 0 \) has only the trivial solution \( x = 0 \).
**Why is this the case?**
The general solution to \( Ax = b \) can be expressed as \( x = x_p + x_h \), where:
- \( x_p \) is a particular solution to \( Ax = b \).
- \( x_h \) is the general solution to the homogeneous equation \( Ax = 0 \).
If \( Ax = 0 \) has only the trivial solution \( x = 0 \), then \( x_h \) must be zero. This means the only solution to \( Ax = b \) is the particular solution \( x_p \), making it unique.
Conversely, if \( Ax = 0 \) has non-trivial solutions, then there are infinitely many solutions to the original equation \( Ax = b \), since you can add any non-zero solution of \( Ax = 0 \) to \( x_p \), resulting in multiple solutions for \( Ax = b \).
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