Definition 1.5.1. Homogeneous System.
A system \(A\vec{x}=\vec{0}\) is called a homogeneous system of equations.
Section 1.5 Homogeneous
We will
- find a relationship between solutions to \(A\vec{x}=\vec{0}\) and \(A\vec{x}=\vec{b}\text{,}\)
- interpret this relationship geometrically, and
- prove it algebraically
By the end of this section you will
- know a relationship between solutions to homogeneous and non-homogeneous systems based on the same matrix, and
- use this to efficiently solve homogeneous and non-homogeneous systems.
Subsection 1.5.1 Discovering Non-Homogeneous Solution Forms
We want to find a connection between solutions to a homogeneous system of equations and the non-homogeneous system with the same matrix.
Definition 1.5.2. Non-Homogeneous System.
A system \(A\vec{x}=\vec{b},\) \(\vec{b} \ne \vec{0}\) is called a non-homogeneous system of equations.
Activity 1.5.1.
\begin{equation*}
A=\left[ \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 3 & 3 \\ 1 & 1 & 2 \end{array} \right], \;\; \vec{b}_1=\left[ \begin{array}{r} 3 \\ 8 \\ 5 \end{array} \right], \;\; \vec{b}_2=\left[ \begin{array}{r} 20 \\ 44 \\ 24 \end{array} \right], \;\; \vec{b}_3=\left[ \begin{array}{r} -7 \\ -13 \\ -6 \end{array} \right]
\end{equation*}
(a)
Find all solutions to \(A\vec{x}=\vec{0}\text{.}\)
Solution.
\(\vec{x} = a[-3,1,1]^T. \)
(b)
Describe these solutions geometrically (we have done this before).
Solution.
Line through origin with slope \([-3,1,1]^T\)
(c)
Find all solutions to \(A\vec{y}=\vec{b}_1\text{.}\)
Solution.
\(\vec{y} = a[-3,1,1]^T+[-7,2,0]^T. \)
(d)
Find all solutions to \(A\vec{y}=\vec{b}_2\text{.}\)
Solution.
\(\vec{y} = a[-3,1,1]^T+[28,-4,0]^T. \)
(e)
Find all solutions to \(A\vec{y}=\vec{b}_3\text{.}\)
Solution.
\(\vec{y} = a[-3,1,1]^T+[-5,-1,0]^T. \)
(f)
For all three non-homogenous solutions what is the same?
Solution.
The \(a[-3,1,1]^T\) part
(g)
For all three non-homogenous solutions what is different? Why might this be different?
Solution.
The vector that is added is different. They are solutions to systems with different right hand sides.
(h)
Describe these solutions geometrically.
Solution.
Lines with slope \([-3,1,1]\) through the point given by the solution (e.g., \([-5,-1,0]^T\) for \(b_3\)).
Now we will connect this idea to math we already know.
Activity 1.5.2.
(a)
In olden days you worked with \(y=mx+b\text{.}\) What affect does \(b\) have on the line?
Solution.
Shifts the line of slope \(m\) up or down by \(b\text{.}\)
(b)
In more recent days (calculus 3) you worked with \(ax+by+cz+d=0\text{.}\) What affect does \(d\) have on the plane?
Solution.
Shifts the plane with normal vector \((a,b,c)\) up or down by \(d\text{.}\)
(c)
With this concept in mind, describe the relationship between \(A\vec{x}=\vec{0}\) and \(A\vec{x}=\vec{p}\text{.}\)
(d)
Use an algebraic property to do something with \(A(\vec{u}+\vec{x})\text{.}\)
Solution.
\(A(\vec{u}+\vec{x}) = A\vec{u}+A\vec{x} \)
(e)
What does \(A(\vec{u}+\vec{x})\) become if \(A\vec{u}=\vec{0}\) and \(A\vec{x}=\vec{b}\text{?}\)
Solution.
\(A\vec{x} \)
Subsection 1.5.2 Proof
Proofs are important in mathematics. Sometimes they explain why something works. This is one of those cases. They also, always, provide a certainty that our conclusions are accurated based on the set of assumptions (axioms) with which we are working. This is important to mathematical culture.
Given
\begin{align*}
A\vec{u} & = \vec{0}.\\
A\vec{x} & = \vec{b}.\\
A(\vec{u}+\vec{x}) & =\\
A\vec{u}+A\vec{x} & =\\
\vec{0}+\vec{b} & = \vec{b}.
\end{align*}
