Solving Systems of Equations with Whole Numbers

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SUMMARY

This discussion addresses the solvability of systems of equations with whole numbers. It establishes that for a system represented as Ax=b, if the determinant det(A) equals -1 and both A and b consist of whole numbers, a unique solution exists with whole number components. Additionally, if det(A) equals 2 and A contains whole numbers while b consists of even numbers, a unique solution also exists with whole number components. The solution can be expressed as x=A-1b, where A-1 is derived using the formula (1/det A)adj A.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically determinants and matrix inverses.
  • Familiarity with the properties of whole numbers and integers.
  • Knowledge of the adjugate matrix and its role in finding inverses.
  • Basic skills in solving systems of linear equations.
NEXT STEPS
  • Study the properties of determinants in linear algebra.
  • Learn about the adjugate matrix and its calculation methods.
  • Explore integer solutions in linear systems, particularly in relation to whole numbers.
  • Investigate the implications of matrix rank on the existence of solutions in systems of equations.
USEFUL FOR

Students and professionals in mathematics, particularly those focused on linear algebra, educators teaching systems of equations, and anyone interested in the properties of integer solutions in mathematical systems.

Yankel
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Hello all

I have a couple of short questions, both similar, which I do not know how to even start, and I could use some help.

1) Show that if in the system Ax=b, det(A)=-1, and all the members of A are whole numbers (belong to Z), and all the members of b are whole numbers (belong to Z), then a single solution exists, and it's members are also whole numbers.

2) Show that if in the system Ax=b, det(A)=2, and all the members of A are whole, and all the members of b are even, then a single solution exists, and all it's members are whole numbers.

Thank you in advance !
 
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The solution is $x=A^{-1}b$. One way to find $A^{-1}$ is $\frac{1}{\det A}\text{adj}\,A$ where $\text{adj}\,A$ is the adjugate of $A$. By construction, if $A$ consists of whole numbers, then so does $\text{adj}\,A$. In the second case, one can factor out 2 from $b$ and cancel it with $\det A$.
 

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