Ch 1.5 - Solution Sets of Linear Systems

1.5 Solution Sets of Linear Systems

Solution sets of linear systems are fundamental in linear algebra and will reappear in various contexts. This section uses vector notation to provide explicit and geometric descriptions of these solution sets.

Vector Equations and System Consistency

• A linear system such as
  A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}, u = \begin{bmatrix} 2 \ 3 \end{bmatrix}, v = \begin{bmatrix} 2 \ 4 \end{bmatrix} leads to vector operations like u+v = \begin{bmatrix} 4 \ 7 \end{bmatrix} and A(u+v) = Au+Av = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 4 \ 7 \end{bmatrix} = \begin{bmatrix} 4 \ 7 \end{bmatrix}

• A system of linear equations is equivalent to a vector equation which expresses the right-hand side vector
  b
  as a linear combination of the columns of the coefficient matrix
  A
  .

• When constructing matrices for specific criteria, using matrices already in reduced echelon form is often straightforward.

• Example of Consistency/Inconsistency using RREF:

  • For Ax = b with augmented matrix RREF: \begin{bmatrix} 1 & 0 & 0 & 2 \ 0 & 1 & 0 & 3 \ 0 & 0 & 0 & 0 \end{bmatrix}

    • This corresponds to a consistent system, meaning Ax = b has solutions.

  • For Ax = c with augmented matrix RREF: \begin{bmatrix} 1 & 0 & 1 & 3 \ 0 & 1 & 1 & 2 \ 0 & 0 & 0 & 1 \end{bmatrix}

    • This corresponds to an inconsistent system (due to the row
      [0 \ 0 \ 0 \ 1]
      ), meaning Ax = c has no solutions.

Homogeneous Linear Systems (Ax = 0)

• A system of linear equations is homogeneous if it can be written in the form Ax = 0

  • Here,
    A
    is an m \times n matrix, and
    0
    is the zero vector in R^m.

• Trivial Solution: Homogeneous systems always have at least one solution, x = 0 (the zero vector in R^n). This is called the trivial solution.

• Nontrivial Solutions: The key question is whether there exists a nonzero vector x that satisfies Ax = 0.

  • Condition for Nontrivial Solutions: The homogeneous equation Ax = 0 has a nontrivial solution if and only if the equation has at least one free variable (referencing the Existence and Uniqueness Theorem from Section 1.2).

Example 1: Homogeneous System with Nontrivial Solutions

• System:
  
  3x1 - 5x2 - 4x3 = 0 \ -3x1 - 2x2 + 4x3 = 0 \
  6x1 - x2 - 8x_3 = 0
  

• Solution Steps:

  1. Form Augmented Matrix [A \ 0] :
     
     \begin{bmatrix}
     3 & -5 & -4 & 0 \
     -3 & -2 & 4 & 0 \
     6 & -1 & -8 & 0
     \end{bmatrix}
     

  2. Row Reduce to Echelon Form:
     
     \begin{bmatrix}
     3 & -5 & -4 & 0 \
     0 & -7 & 0 & 0 \
     0 & 9 & 0 & 0
     \end{bmatrix}
     \sim
     \begin{bmatrix}
     3 & -5 & -4 & 0 \
     0 & -7 & 0 & 0 \
     0 & 0 & 0 & 0
     \end{bmatrix}
     

  • Since x_3 is a free variable, Ax = 0 has nontrivial solutions.

  1. Continue Row Reduction to Reduced Echelon Form:
     
     \begin{bmatrix}
     1 & 0 & -4/3 & 0 \
     0 & 1 & 0 & 0 \
     0 & 0 & 0 & 0
     \end{bmatrix}
     

  2. Solve for Basic Variables:

     • The equations are:
       
       x1 - \frac{4}{3}x3 = 0 \
       x_2 = 0 \
       0 = 0
       

     • This yields x1 = \frac{4}{3}x3, x2 = 0, with x3 free.

  3. General Solution in Vector Form: x = \begin{bmatrix} x1 \ x2 \ x3 \end{bmatrix} = \begin{bmatrix} (4/3)x3 \ 0 \ x3 \end{bmatrix} = x3 \begin{bmatrix} 4/3 \ 0 \ 1 \end{bmatrix}

     • This can be written as x = x_3v, where v = \begin{bmatrix} 4/3 \ 0 \ 1 \end{bmatrix}.

     • Every solution is a scalar multiple of
       v
       .

  • Geometric Interpretation: The solution set is a line through the origin in R^3 (Figure 1).

  • Note: A nontrivial solution can have zero entries, as long as not all entries are zero.

Example 2: Single Linear Equation as Homogeneous System

• System: 10x1 - 3x2 + 2x_3 = 0

• Solution Steps:

  1. Solve for Basic Variable: Solve for x1 in terms of free variables x2 and x3: x1 = 0.3x2 - 0.2x3.

  2. General Solution in Vector Form: x = \begin{bmatrix} x1 \ x2 \ x3 \end{bmatrix} = \begin{bmatrix} 0.3x2 - 0.2x3 \ x2 \ x3 \end{bmatrix} = x2 \begin{bmatrix} 0.3 \ 1 \ 0 \end{bmatrix} + x_3 \begin{bmatrix} -0.2 \ 0 \ 1 \end{bmatrix}

     • Let u = \begin{bmatrix} 0.3 \ 1 \ 0 \end{bmatrix} and v = \begin{bmatrix} -0.2 \ 0 \ 1 \end{bmatrix}.

     • The solution set is Span\left{ u, v \right}.

  • Geometric Interpretation: Since neither
    u
    nor
    v
    is a scalar multiple of the other, the solution set is a plane through the origin (Figure 2).

• Generalization for Homogeneous Systems: The solution set of Ax = 0 can always be expressed as Span\left{ \mathbf{v1}, \ldots, \mathbf{vp} \right} for suitable vectors.

  • If only the zero vector is a solution, the set is Span\left{ 0 \right}.

  • If there's one free variable, the solution set is a line through the origin.

  • If there are two or more free variables, the solution set is a plane through the origin (or higher-dimensional subspace).

Parametric Vector Form

• Definition: An explicit description of a solution set using vectors (like x = x3v or x = x2u + x_3v) is called the parametric vector form.

• This form emphasizes that parameters (e.g., x3, x2, x_3 in the examples, or s, t \in R) vary over all real numbers.

  • For a line: x = tv (where
    t \in R
    ).

  • For a plane: x = su + tv (where
    s, t \in R
    ).

• The original equation (e.g., 10x1 - 3x2 + 2x_3 = 0) is an implicit description, while the parametric vector form is an explicit description.

Solutions of Nonhomogeneous Systems (Ax = b)

• When a nonhomogeneous system has multiple solutions, its general solution can be written as one particular vector plus an arbitrary linear combination of vectors that satisfy the corresponding homogeneous system (Ax = 0).

Example 3: Nonhomogeneous System Solution

• System: Ax = b where
  A = \begin{bmatrix} 3 & -5 & -4 \ -3 & -2 & 4 \ 6 & -1 & -8 \end{bmatrix} and b = \begin{bmatrix} -7 \ 7 \ -4 \end{bmatrix}. (Matrix A is from Example 1).

• Solution Steps:

  1. Form Augmented Matrix [A \ b] :
     
     \begin{bmatrix}
     3 & -5 & -4 & -7 \
     -3 & -2 & 4 & 7 \
     6 & -1 & -8 & -4
     \end{bmatrix}
     

  2. Row Operations to Reduced Echelon Form:
     
     \begin{bmatrix}
     1 & 0 & -4/3 & -1 \
     0 & 1 & 0 & 2 \
     0 & 0 & 0 & 0
     \end{bmatrix}
     

  3. Solve for Basic Variables:

     • The equations are:
       
       x1 - \frac{4}{3}x3 = -1 \
       x_2 = 2 \
       0 = 0
       

     • This yields x1 = -1 + \frac{4}{3}x3, x2 = 2, with x3 free.

  4. General Solution in Vector Form (Parametric Vector Form): x = \begin{bmatrix} x1 \ x2 \ x3 \end{bmatrix} = \begin{bmatrix} -1 + (4/3)x3 \ 2 \ x3 \end{bmatrix} = \begin{bmatrix} -1 \ 2 \ 0 \end{bmatrix} + x3 \begin{bmatrix} 4/3 \ 0 \ 1 \end{bmatrix}

     • This can be written as x = p + x_3v (or x = p + tv where t \in R ).

       • 
         p = \begin{bmatrix} -1 \ 2 \ 0 \end{bmatrix}
         is a particular solution (obtained when x_3 = 0 or t=0).

       • 
         v = \begin{bmatrix} 4/3 \ 0 \ 1 \end{bmatrix}
         is the same vector from the solution of the corresponding homogeneous system (Ax=0) in Example 1.

• Geometric Interpretation: Translation

  • Vector addition can be seen as a translation. Adding a vector
    p
    to another vector
    v
    translates
    v
    to v + p. (Figure 3).

  • If a line L (e.g., the solution set of Ax=0) is translated by a vector
    p
    , the result is a parallel line (Figure 4).

  • Therefore, the solution set of Ax = b is a line through
    p
    parallel to the solution set of Ax = 0 (Figure 5).

Theorem 6: Solution Set of Ax=b

• Suppose the equation Ax = b is consistent for some given
  b
  , and let
  p
  be a particular solution.

• Then, the solution set of Ax = b is the set of all vectors of the form w = p + vh, where vh
  is any solution of the homogeneous equation Ax = 0.

• Implication: The solution set of Ax = b is obtained by translating the solution set of Ax = 0 using any particular solution p of Ax = b for the translation (Figure 6).

  • If there are two free variables, the solution set is a plane not passing through the origin.

• Warning: This theorem applies only if Ax=b is consistent and has at least one solution
  p
  . If Ax = b has no solution, the solution set is empty.

Algorithm for Writing a Solution Set in Parametric Vector Form

1. Row reduce the augmented matrix to reduced echelon form.

2. Express each basic variable in terms of any free variables appearing in an equation.

3. Write a typical solution x as a vector whose entries depend on the free variables (if any).

4. Decompose x into a linear combination of vectors with numeric entries, using the free variables as parameters.

Verifying Solutions ("Reasonable Answers")

• For Homogeneous Equation (Ax = 0):

  • Multiply the matrix A by each vector in your solution (e.g., the
    v
    vector in x = x_3v).

  • The result should be the zero vector.

  • Example: If
    A = \begin{bmatrix} -2 & 1 & 2 \ -1 & 2 & 5 \ 1 & 1 & 3 \end{bmatrix} and solution is x_3 \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix}, verify
    A \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 2 \ -1 & 2 & 5 \ 1 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} (-2)(1) + (1)(1) + (2)(-1) \ (-1)(1) + (2)(1) + (5)(-1) \ (1)(1) + (1)(1) + (3)(-1) \end{bmatrix} = \begin{bmatrix} -2+1-2 \ -1+2-5 \ 1+1-3 \end{bmatrix} = \begin{bmatrix} -3 \ -4 \ -1 \end{bmatrix}. The example provided verification for a vector that yields
    [0 \ 0 \ 0]
    , illustrating the correct check. If your solution vector is
    u = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix}
    , then
    Au = 0
    must hold.

• For Nonhomogeneous Equation (Ax = b):

  • Multiply the matrix A by the first vector in your solution (the particular solution
    p
    ).

  • The product should be
    b
    .

  • Multiply the matrix A by the remaining vectors (those that are part of the homogeneous solution).

  • The product of A with these vectors should be the zero vector.

  • Example: If x = \begin{bmatrix} -3 \ 1 \ 1 \end{bmatrix} + x3 \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} + x4 \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} is a solution to Ax=b, where A = \begin{bmatrix} -2 & 1 & 2 \ -1 & 2 & 5 \ 1 & 1 & 3 \end{bmatrix} and b = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}. Using prior calculations for A\begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} and A\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 6 \ 5 \end{bmatrix}, verify the particular solution:
    
    A \begin{bmatrix} -3 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 2 \ -1 & 2 & 5 \ 1 & 1 & 3 \end{bmatrix} \begin{bmatrix} -3 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} (-2)(-3) + (1)(1) + (2)(1) \ (-1)(-3) + (2)(1) + (5)(1) \ (1)(-3) + (1)(1) + (3)(1) \end{bmatrix} = \begin{bmatrix} 6+1+2 \ 3+2+5 \ -3+1+3 \end{bmatrix} = \begin{bmatrix} 9 \ 10 \ 1 \end{bmatrix}. The example implies that this should equal
    b
    , which it currently doesn't match the given
    b
    . The example given for verification on page 6 is more complex, but the general principle is correct: the portion multiplied by A that is NOT part of the homogeneous solution should yield
    b
    , and the portions of the homogeneous solution still multiplied by A should yield
    0
    .