HOMOGENEOUS SYSTEMS OF LINEAR EQUATIONS

Def'n: A system of linear equations is homogeneous if all the constant terms (b's) are 0.

For a homogeneous system, exactly one of the following is true:

1) the system had only the trivial solution.(all variables = 0)

2) the system has infinitely many nontrivial solutions in addition to the trivial solution.
This is always true when there are more variables than equations in the system --
when there are more columns than rows in the coefficient matrix.
In such a case, we use parameters to solve the system.

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Example 1

Solve this system by elimination.

Solution

The augmented matrix for this system is

, which we reduce to row-echelon form to get

, so we know that

It is obvious then that x ₄ = 0.
If we had a solution for x ₅ in the 2nd equation, we could find x ₃ , so we
set x ₅ = t, which makes x ₃ = – t.
Similarly, if we had a solution for x ₂ in the first equation, we could find x ₁ , so we
set x ₂ = s, which makes x ₁ = – s – t.
The general solution for this system is:

x 1 = – s – t

x2 = s

x 3 = – t

x 4 = 0

x 5 = t

t ` R

Now, when we choose any real number value for t and s,
we will have values for all 5 unknown variables. If we choose
both t and s = 0, we have the trivial solution. All other real values
for t and s will produce the infinite nontrivial solutions.

Since one of the equations was eliminated, (row of zeros at bottom), we're left
with 3 equations in 5 variables -- so we need 2 parameters to solve. For 1 equation
in 3 variables we'd need 3 – 1 = 2 parameters like this:

Example 2: x₁ + 3x₂ – x₃ = 0 is solved with 2 parameters.
So, we let x₂ = t, and x₃ = s, which makes x₁ = –3t + s; t and s are elements of R.

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Practice

1) Solve these homogeneous systems.

a)	b)
c)	d)

2) For which values of k does this system have nontrivial solutions?

3) Consider the system of equations

Discuss the relative positions of the lines in this system when:
a) the system has only the trivial solution
b) the system has nontrivial solutions.

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Solutions

1) a) To set up the augmented matrix:

first, switch row2 with row1, and switch row3 with row2 to get:

, now (row3 – 2row1)/3 and ½(row3 + row2) to get:

, so the system has only the trivial solution x₁ = x₂ = x₃ = 0.

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b) The augmented matrix .

Since we have 4 variables and 2 equations, we need 2 parameters:

Let x₄ = t and let x₃ = 4s, which means that x₂ = – t – s and x₁ = – s.

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c) To create the augmented matrix, switch the 1st and 2nd equations,
then use ½ (eq. 1) for row2 to get:

, now use row1 to eliminate entries in the first column.

step 2: (row3 – 2row1) and (row4 + 2row1)

step 3: (row3 – 3row2) and (row4 – row2)

Now we've got , so we add 10 row3 to row 4 to get zeros in the last row.

When we set y = s, we get

w = s

x = – s

y = s

z = 0

s R

d) To set up the augmented matrix: first, switch equation3 with equation1 to get:

, now (row2 + row1) and (row3 – 2row1) to get:

, now add row3 to row 2 to get ,

and when we multiply row2 by 1/3 and row3 by – 1/10, we get the reduced matrix

, which means the system has only the trivial solution x₁ = x₂ = x₃ = 0.

2) If we use the fact that the determinant must = 0 to have nontrivial solutions, we get:
(k – 3)² – 1 = 0 which means that k² – 6k + 8 = 0, so (k – 4)(k – 2) = 0
If k = 4 or k = 2, there will be nontrivial solutions.

Using row reduction, we will switch the 2 rows to get:

, so we'll take row2 – (k – 3)row1 to get:

and when we set 1 – (k – 3)² = 0 we get the same results.

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3) a) If the system has only the trivial solution, at least 2 of the lines are distinct and they
all intersect at a single point, the origin.

b) For the system to have nontrivial solutions, all 3 lines must be coincident (the same line).

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