Homogeneous Linear Systems
A particular case of the linear system defined in the previous lesson is when the constants $b_i$ are all equal to zero. In this case, we say that the linear system is homogeneous.
of the form
\begin{eqnarray}\label{homogeneous-linear-system}
\begin{array}{ccccccccccc}
a_{11}x_1 & +& a_{12}x_2& +& a_{13}x_3& +& \cdots & +&a_{1n}x_n&=&0\\
a_{21}x_1 & +& a_{22}x_2& +& a_{23}x_3& +& \cdots & +&a_{2n}x_n&=&0\\
a_{31}x_1 & +& a_{32}x_2& +& a_{33}x_3& +& \cdots & +&a_{3n}x_n&=&0\\
& \vdots & & \vdots & & \vdots & & \vdots & &\vdots &\\
a_{m1}x_1 & +& a_{m2}x_2& +& a_{m3}x_3& +& \cdots & +&a_{mn}x_n&=&0
\end{array}
\end{eqnarray}
is called a homogeneous linear system.
If we set $x_1 = 0,$ $x_2 = 0,\ldots, x_n = 0 $, we see that the system is satisfied. Hence $(0,0,\cdots,0)$ is always a solution of the homogeneous linear system \eqref{homogeneous-linear-system}. Therefore the homogeneous linear system \eqref{homogeneous-linear-system} is always consistent. The solution $(0,0,\cdots,0)$ is called the trivial solution; if there are other solutions, they are called nontrivial solutions.
Because a homogeneous linear system always has a trivial solution, there are only
two possibilities for its solutions:
- The system has only a trivial solution.
- The system has infinitely many solutions in addition to the trivial solution.
of its augmented matrix has $r$ nonzero rows, then the system has $n-r$ free variables.
The idea is to use the Free Variable Theorem for Homogeneous Systems. It is given that the number of unknowns is $n=9$ and that the RREF has $r=4$ nonzero rows. It follows by Theorem ?? that there are $9-4=5$ free variables.
It is given that the number of unknowns is $n=8$ and that the RREF has $r=2$ nonzero rows. It follows by Theorem ?? that there are $8-2=6$ free variables.
Consider a homogeneous linear system $A\mathrm{X}=0$, where $A$ is a non-zero $4\times 5$ matrix. What are the maximum and minimum parameters in the general solution to this system?
The number of variables is $n=5$, and the number of equations is $m=4$. Let $r$ be the rank of $A$ ($=$ the number of leading $1’\mathrm{s}$ in the RREF of $A$).
Suppose we have a homogeneous system of $m$ equations in $n$ variables $x_1, x_2,\ldots,x_n$. If $n>m$, that is, if there are more variables than equations, the system has infinitely many solutions. It is also possible, but not required, to have a nontrivial solution if $n=m$ and $n<m$.
infinitely many solutions.
Consider the linear system
\begin{eqnarray*}
\begin{array}{rrrrrrrrr}
x_1&+&2x_2&+&x_3& & =&0\\
2x_1&-&2x_2& +&3x_3& &=&0\\
\end{array}
\end{eqnarray*}
- Without solving the linear system, could you predict the number of solutions?
- Solve the linear system by Gauss-Jordan elimination.
- This is a homogeneous linear system of $2$ equations and $3$ unknowns. Since there are more unknowns than equations, there are infinitely many solutions.
- First, we find the RREF of the corresponding augmented matrix. The augmented is
$$\left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
2 & -2 & 3 & 0
\end{array}\right].$$
We apply the following elementary row operations to the augmented matrix to obtain the RREF.
\begin{eqnarray*}
& & \left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
2 & -2 & 3 & 0
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_2\to R_2-2R_1}\\
R_3\to R_3-R_1\\
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
0 & -6 & 1 & 0
\end{array}\right]\\
& &\begin{array}{r}
\\
\\
\underrightarrow{R_2\to -\frac{1}{6}R_2}\\
{}\\
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
0 & 1 & -\frac{1}{6} & 0
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_1\to R_1-2R_2}\\
{}\\
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & 0 & \frac{4}{3} & 0 \\
0 & 1 & -\frac{1}{6} & 0
\end{array}\right]
\end{eqnarray*}
The linear system corresponding to the RREF above is
\begin{eqnarray*}
\begin{array}{rrrrrrr}
x& & & +& \frac{4}{3}z &=&0\\
& & y&-&\frac{1}{6}z&=&0
\end{array}
\end{eqnarray*}The leading variables are $x$ and $y$, and the free variable is $z$. We can set the free variable $z$ as a parameter, say $z=t$, and solve for $x$ and $y$ in terms of the parameter $t$. Hence the parametric solutions are given by
\begin{eqnarray*}
x&=&-\frac{4t}{3}\\
y&=&\frac{t}{6}\\
z&=&t,\qquad t\in\mathbb{R}.
\end{eqnarray*}
Problem 1. Consider the following system of linear equations:
\begin{eqnarray*}
\begin{array}{rrrrrrrrr}
4x_1&-&x_2&-&x_3& & & =&160\\
-x_1&+&4x_2& & &-&x_4&=&140\\
-x_1& & & +&4x_3 &-&x_4&=&60\\
&-&x_2& -&x_3 &+&4x_4&=&40
\end{array}
\end{eqnarray*}
- Write the system as an augmented matrix and find its RREF.
- Solve the system.
Solution of Problem 1.
- The augmented matrix is
$$
\left[\begin{array}{rrrr|r}
4 & -1 & -1 & 0 & 160 \\
-1 & 4 & 0 & -1 & 140 \\
-1 & 0 & 4 & -1 & 60 \\
0 & -1 & -1 & 4 & 40 \\
\end{array}\right]$$
Next, we find the RREF using elementary row operations. We have\begin{eqnarray*}
& & \left[\begin{array}{rrrr|r}
4 & -1 & -1 & 0 & 160 \\
-1 & 4 & 0 & -1 & 140 \\
-1 & 0 & 4 & -1 & 60 \\
0 & -1 & -1 & 4 & 40 \\
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_1\leftrightarrow R_2}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
-1 & 4 & 0 & -1 & 140 \\
4 & -1 & -1 & 0 & 160 \\
-1 & 0 & 4 & -1 & 60 \\
0 & -1 & -1 & 4 & 40 \\
\end{array}\right]\\
& &\begin{array}{r}
\\
\\
\underrightarrow{R_1\to -R_1}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140 \\
4 & -1 & -1 & 0 & 160 \\
-1 & 0 & 4 & -1 & 60 \\
0 & -1 & -1 & 4 & 40 \\
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_2\to R_2-4R_1}\\
R_3\to R_3+R_1\\
{}
\end{array}\\
&&
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & 15 & -1 & -4 & 720 \\
0 & -4 & 4 & 0 & -80\\
0 & -1 & -1 & 4 & 40 \\
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_2\leftrightarrow R_4}\\
{}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & -1 & -1 & 4 & 40 \\
0 & -4 & 4 & 0 & -80\\
0 & 15 & -1 & -4 & 720\\
\end{array}\right]\\
&&\begin{array}{r}
\\
\\
\underrightarrow{R_2\to- R_2}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & 1 & 1 & -4 & -40 \\
0 & -4 & 4 & 0 & -80\\
0 & 15 & -1 & -4 & 720\\
\end{array}\right]\begin{array}{r}
\\
\\
\underrightarrow{R_3\to R_3+4 R_2}\\
R_4\to R_4-15R_2
{}
\end{array}\\
&&
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & 1 & 1 & -4 & -40 \\
0 & 0 & 8 & -16 & -240\\
0 & 0 & -16 & 56 & 1320\\
\end{array}\right]\begin{array}{r}
\\
\\
\underrightarrow{R_3\to \frac{1}{8} R_3}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & 1 & 1 & -4 & -40 \\
0 & 0 & 1 & -2 & -30\\
0 & 0 & -16 & 56 & 1320\\
\end{array}\right]\\
&&
\begin{array}{r}
\\
\\
\underrightarrow{R_4\to R_4+16R_3}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & 1 & 1 & -4 & -40 \\
0 & 0 & 1 & -2 & -30\\
0 & 0 & 0 & 24 & 840\\
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_4\to\frac{1}{4} R_4}\\
{}
\end{array}\\
&&\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 1 & -140\\
0 & 1 & 1 & -4 & -40 \\
0 & 0 & 1 & -2 & -30\\
0 & 0 & 0 & 1 & 35\\
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_3\to R_3+2 R_4}\\
R_2\to R_2+4R_4\\
R_1\to R_1-R_4
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 0 & -175\\
0 & 1 & 1 & 0 & 100 \\
0 & 0 & 1 & 0 & 40\\
0 & 0 & 0 & 1 & 35\\
\end{array}\right]\\
\end{eqnarray*}
\begin{eqnarray*}
&&\begin{array}{r}
\\
\\
\underrightarrow{R_2\to R_2- R_3}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & -4 & 0 & 0 & -175\\
0 & 1 & 0 & 0 & 60 \\
0 & 0 & 1 & 0 & 40\\
0 & 0 & 0 & 1 & 35\\
\end{array}\right]\begin{array}{r}
\\
\\
\underrightarrow{R_1\to R_1+4 R_2}\\
{}
\end{array}\\
&&
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & 0 & 65\\
0 & 1 & 0 & 0 & 60 \\
0 & 0 & 1 & 0 & 40\\
0 & 0 & 0 & 1 & 35\\
\end{array}\right]\qquad RREF
\end{eqnarray*}
(b)
Going back to the linear system, we obtain the following:
\begin{eqnarray*}
x_1&=&65\\
x_2&=&60\\
x_3&=&40\\
x_4&=&35
\end{eqnarray*}
Problem 2. Consider the following system of linear equations:
\begin{eqnarray*}
\begin{array}{rrrrrrrrr}
x_1&-&x_2&+&4x_3& =&7\\
x_1&+&3x_2&- &6x_3&=&-13\\
2x_1&+&2x_2& -&2x_3 &=&-10
\end{array}
\end{eqnarray*}
- Write the system as an augmented matrix and find its RREF.
- Solve the system.
Solution of Problem 2.
- The augmented matrix is
$$
\left[\begin{array}{rrr|r}
1 & -1 & 4 & 7 \\
1 & 3 & -6 & -13 \\
2 & 2 & -2 & -10
\end{array}\right]$$
Next, we find the RREF using elementary row operations. We have\begin{eqnarray*}
& &\left[\begin{array}{rrr|r}
1 & -1 & 4 & 7 \\
1 & 3 & -6 & -13 \\
2 & 2 & -2 & -10
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_2\leftrightarrow R_2-R_1}\\
R_3\to R_3-2R_1
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & -1 & 4 & 7 \\
0 & 4 & -10 & -20 \\
0 & 4 & -10 & -24
\end{array}\right]\\
& &\begin{array}{r}
\\
\\
\underrightarrow{R_3\to R_3-R_2}\\
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & -1 & 4 & 7 \\
0 & 4 & -10 & -20 \\
0 & 0 & 0 & -4
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_2\to \frac{1}{4}R_2}\\
R_3\to -\frac{1}{4}R_3\\
{}
\end{array}\\
&&
\left[\begin{array}{rrr|r}
1 & -1 & 4 & 7 \\
0 & 1 & -5/2 & -5 \\
0 & 0 & 0 & 1
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_2\to R_2+5R_3}\\
R_1\to R_1-7R_3\\
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & -1 & 4 & 0 \\
0 & 1 & -5/2 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]\\
&&\begin{array}{r}
\\
\\
\underrightarrow{R_1\to R_1+R_2}\\
{}
\end{array}
\left[\begin{array}{rrr|r}
1 & 0 & 3/2 & 0 \\
0 & 1 & -5/2 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]\qquad\mbox{RREF}
\end{eqnarray*} - Going to the linear system, we obtain
\begin{eqnarray*}
x_1+\frac{3}{2}x_3&=&0\\
x_2-\frac{5}{2}x_3&=&0\\
0&=&1.
\end{eqnarray*}
The last equation, $0=1$, gives a contradiction. Therefore is no solution.
Problem 3. Consider the following system of linear equations:
\begin{eqnarray*}
\begin{array}{rrrrrrrrr}
x_1&+&2x_2&-&3x_3& +& x_4& =&13\\
2x_1&+&x_2&- &4x_3 &-&2x_4&=&13\\
3x_1& -&2x_2 & +&x_3 &+&x_4&=&-5\\
6x_1&+&x_2& -&6x_3 &&&=&21
\end{array}
\end{eqnarray*}
- Write the system as an augmented matrix and find its RREF.
- Solve the system.
Solution of Problem 3.
- The augmented matrix is
$$
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
2 & 1 & -4 & -2 & 13 \\
3 & -2 & 1 & 1 & -5 \\
6 & 1 & -6 & 0 & 21
\end{array}\right]$$
Next, we find the RREF using elementary row operations. We have\begin{eqnarray*}
& & \left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
2 & 1 & -4 & -2 & 13 \\
3 & -2 & 1 & 1 & -5 \\
6 & 1 & -6 & 0 & 21
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_2\leftrightarrow R_2-2R_1}\\
R_3\to R_3-3R_1\\
R_3\to R_3-6R_1
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
0 & -3 & 2 & -4 & -13 \\
0 & -8 & 10 & -2 & -44 \\
0 & -11 & 12 & -6 & -57\\
\end{array}\right]\\
& &\begin{array}{r}
\\
\\
\underrightarrow{R_3\to R_3-3R_2}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
0 & -3 & 2 & -4 & -13 \\
0 & 1 & 4 & 10 & -5 \\
0 & -11 & 12 & -6 & -57\\
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_2\leftrightarrow R_3}\\
{}
\end{array}\\
&&
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
0 & 1 & 4 & 10 & -5 \\
0 & -3 & 2 & -4 & -13 \\
0 & -11 & 12 & -6 & -57\\
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_3\to R_3+3R_2}\\
R_4\to R_4+11R_2\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
0 & 1 & 4 & 10 & -5 \\
0 & 0 & 14 & 26 & -28 \\
0 & 0 & 56 & 104 & -112
\end{array}\right]\\
&&\begin{array}{r}
\\
\\
\underrightarrow{R_4\to R_4-4R_3}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
0 & 1 & 4 & 10 & -5 \\
0 & 0 & 14 & 26 & -28 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_3\to \frac{1}{14} R_3}\\
{}
\end{array}\\
&&
\left[\begin{array}{rrrr|r}
1 & 2 & -3 & 1 & 13 \\
0 & 1 & 4 & 10 & -5 \\
0 & 0 & 1 & \frac{13}{7} & -2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]\begin{array}{r}
\\
\\
\underrightarrow{R_2\to R_2-4 R_3}\\
R_1\to R_1+3R_3
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & 2 & 0 & \frac{46}{7} & 7 \\
0 & 1 & 0 & \frac{18}{7} & 3 \\
0 & 0 & 1 & \frac{13}{7} & -2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]\\
&&\begin{array}{r}
\\
\\
\underrightarrow{R_1\to R_1-2 R_2}\\
{}
\end{array}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & \frac{10}{7} & 1 \\
0 & 1 & 0 & \frac{18}{7} & 3 \\
0 & 0 & 1 & \frac{13}{7} & -2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
\qquad\mbox{RREF}
\end{eqnarray*} - Going to the linear system, we obtain
\begin{eqnarray*}
x_1+ \frac{10}{7}x_4&=&1\\
x_2+\frac{18}{7}x_4&=&0\\
x_3+\frac{13}{7}x_4&=&-2.
\end{eqnarray*}
The leading variables are $x_1$, $x_2$, and $x_3$. Only $x_4$ is a free variable. Now we set the free variable as a parameter, say $x=t$, and we solve for $x_1$, $x_2$ and $x_3$.
It follows that the parametric solutions are
\begin{eqnarray*}
x_1&=&1-\frac{10}{7}t\\
x_2&=&-\frac{18}{7}t\\
x_3&=&-2-\frac{13}{7}t\\
x_4&=&t \qquad t\in\mathbb{R}
\end{eqnarray*}
Problem 4. Consider a system of linear equations:
\begin{eqnarray*}
\begin{array}{ccccccccccc}
a_{11}x_1 & +& a_{12}x_2& +& a_{13}x_3& +& \cdots & +&a_{1n}x_n&=&b_1\\
a_{21}x_1 & +& a_{22}x_2& +& a_{23}x_3& +& \cdots & +&a_{2n}x_n&=&b_2\\
a_{31}x_1 & +& a_{32}x_2& +& a_{33}x_3& +& \cdots & +&a_{3n}x_n&=&b_3\\
& \vdots & & \vdots & & \vdots & & \vdots & &\vdots &\\
a_{m1}x_1 & +& a_{m2}x_2& +& a_{m3}x_3& +& \cdots & +&a_{mn}x_n&=&b_m
\end{array}
\end{eqnarray*}
- If $b_1 = b_2 = b_3 = \cdots = b_m = 0$, then the system is homogeneous. Can you find a solution to a homogeneous system without solving it? If so, what is it? (Such a solution is called the trivial solution)
- How many solutions can a homogenous system of linear equations have?
Solution of Problem 4.
- $(0,0,\cdots,0)$ is always a solution of the homogeneous linear system.
- For a homogeneous linear system, we have only two possibilities:
either it has exactly one solution (in this case, $(0,0,\cdot,0)$ is the only solution), or it has infinitely many solutions (of course, including the trivial solution).
