Summary: Possibilities for the Solution Set of a System of Linear Equations

Problem 288

In this post, we summarize theorems about the possibilities for the solution set of a system of linear equations and solve the following problems.

Determine all possibilities for the solution set of the system of linear equations described below.

(a) A homogeneous system of $3$ equations in $5$ unknowns.

(b) A homogeneous system of $5$ equations in $4$ unknowns.

(c) A system of $5$ equations in $4$ unknowns.

(d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution.

(e) A homogeneous system of $4$ equations in $4$ unknowns.

(f) A homogeneous system of $3$ equations in $4$ unknowns.

(g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution.

(h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$.

(i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$.

(j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$.

the possibilities for the solution set of a system of linear equations

An $m\times n$ system of linear equations is
\begin{align*} \tag{*}
a_{1 1} x_1+a_{1 2}x_2+\cdots+a_{1 n}x_n& =b_1 \\
a_{2 1} x_1+a_{2 2}x_2+\cdots+a_{2 n}x_n& =b_2 \\
a_{3 1} x_1+a_{3 2}x_2+\cdots+a_{3 n}x_n& =b_3 \\
a_{m 1} x_1+a_{m 2}x_2+\cdots+a_{m n}x_n& =b_m,
\end{align*}
where $x_1, x_2, \dots, x_n$ are unknowns (variables) and $a_{i j}, b_k$ are numbers.
Thus an $m\times n$ system of linear equations consists of $m$ equations and $n$ unknowns $x_1, x_2, \dots, x_n$.
A system of linear equations is called homogeneous if the constants $b_1, b_2, \dots, b_m$ are all zero. Namely, a homogeneous system is
\begin{align*}
a_{1 1} x_1+a_{1 2}x_2+\cdots+a_{1 n}x_n& =0 \\
a_{2 1} x_1+a_{2 2}x_2+\cdots+a_{2 n}x_n& =0 \\
a_{3 1} x_1+a_{3 2}x_2+\cdots+a_{3 n}x_n& =0 \\
a_{m 1} x_1+a_{m 2}x_2+\cdots+a_{m n}x_n& =0.
\end{align*}
A solution of the system (*) is a sequence of numbers $s_1, s_2, \dots, s_n$ such that the substitution $x_1=s_1, x_2=s_2, \dots, x_n=s_n$ satisfies all the $m$ equations in the system (*).
We sometimes use the vector notation and say
$\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}=\begin{bmatrix} s_1 \\ s_2 \\ \vdots \\ s_n \end{bmatrix}$ is a solution of the system.
For example, every homogeneous system has the zero solution $x_1=0, x_2=0, \dots, x_n=0$, or
$\mathbf{x}=\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.$ Here we summarize several theorems concerning with the possibilities for the number of solutions of a system of linear equations.

We say a system is consistent if the system has at least one solution.
A system is called inconsistent if the system has no solutions at all.

Theorem 1. For a given system of linear equations, there are three possibilities for the solution set of the system: No solution (inconsistent), a unique solution, or infinitely many solutions.

Thus, for example, if we find two distinct solutions for a system, then it follows from the theorem that there are infinitely many solutions for the system.

Next, since a homogeneous system has the zero solution, it is always consistent. Thus:

Theorem 2. The possibilities for the solution set of a homogeneous system is either a unique solution or infinitely many solutions.

Let us refine these theorems. Suppose that an $m\times n$ system of linear equations is given. That is, there are $m$ linear equations and $n$ unknowns.

Theorem 3. If $m < n$, then the system is either inconsistent or it has infinitely many solutions.

Thus, there are only two possibilities when $m < n$: No solution or infinitely many solutions.

Theorem 4. Consider $m\times n$ homogeneous system of linear equations. Then the system has always infinitely many solutions.

This is obtained by noting that a homogeneous system always has the zero solution, hence consistent. By the previous theorem, the only possibility is infinitely many solutions.

Summary 1

Summary 1: The number of solutions of a system of linear equations is one of $0$, $1$, or $\infty$. If there are more unknowns ($n$) than the number of equations ($m$), then the number of solutions of the system is either $0$ or $\infty$. If a system is homogeneous, then it has the zero solution and thus a homogeneous system is always consistent.

The case $m \geq n$?

What happens when $m \geq n$?
In general, when the number of equations is greater than or equal to the number of unknowns, we cannot narrow down the possibilities.
For a given system (*), let $A$ be the coefficient matrix and let $\mathbf{b}$ be the constant term vector. Then we define the rank of the system to be the rank of the augmented matrix $[A\mid \mathbf{b}]$.
Recall that the rank is defined as follows. We first reduce the matrix $[A\mid \mathbf{b}]$ to a matrix $[A’\mid \mathbf{b’}]$ in (reduced) row echelon form by elementary row operations.
Then the rank of $[A\mid \mathbf{b}]$ is the number of nonzero rows in the matrix $[A’\mid \mathbf{b’}]$.

Theorem 5. Consider an $m\times n$ system of linear equations. Suppose that it is consistent. Then the rank $r$ of the system satisfies $r\leq n$. Also, the system has $n-r$ free variables.

A free variable means an unknown that can be assigned arbitrary values. It follows that if a system has a free variable, then there are infinitely many solutions.

Caution: the theorem assumes that a given system is consistent.

Summary 2

Summary 2: Suppose that an $m\times n$ system of linear equations is consistent and let $r$ be the rank of the system. If $n=r$, then the system has a unique solution. If $n>r$, then the system has infinitely many solutions.

Problems and solutions

Determine all possibilities for the solution set of the system of linear equations described below.
(a) A homogeneous system of $3$ equations in $5$ unknowns.

(b) A homogeneous system of $5$ equations in $4$ unknowns.

(c) A system of $5$ equations in $4$ unknowns.

(d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution.

(e) A homogeneous system of $4$ equations in $4$ unknowns.

(f) A homogeneous system of $3$ equations in $4$ unknowns.

(g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution.

(h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$.

(i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$.

(j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$.

Solutions.

In the solution, $m$ denotes the number of equations and $n$ denotes the number of unknowns in the given system.

(a) A homogeneous system of $3$ equations in $5$ unknowns.

Since the system is homogeneous, it has the zero solution, hence consistent. Since there are more unknowns than equations, there are infinitely many solutions.

(b) A homogeneous system of $5$ equations in $4$ unknowns.

Since the system is homogeneous, it has the zero solution. Since there are more equations than unknowns, we cannot determine further.
Thus the possibilities are either a unique solution or infinitely many solution.
(If the rank $r$ of the system is $4$, then a unique solution. If $r<4$, then there are infinitely many solutions.)

(c) A system of $5$ equations in $4$ unknowns.

Since $m > n$, we can only say that the possibilities are no solution, a unique solution, or infinitely many solution.

(d) A system of $2$ equations in $3$ unknowns that has $x_1=1, x_2=-5, x_3=0$ as a solution.

Since $m < n$, the system is either inconsistent or has infinitely many solutions. Since $x_1=1, x_2=-5, x_3=0$ is a solution of the system, the system is not inconsistent. Thus the only possibility is infinitely many solutions.

(e) A homogeneous system of $4$ equations in $4$ unknowns.

Since $m=n$, this tells nothing. But since the system is homogeneous it has the zero solution, hence consistent. The possibilities are either a unique solution or infinitely many solutions.

(f) A homogeneous system of $3$ equations in $4$ unknowns.

Since $m < n$, the system has no solution or infinitely many solutions. But a homogeneous system is always consistent. Thus, the only possibility is infinitely many solutions.

(g) A homogeneous system that has $x_1=3, x_2=-2, x_3=1$ as a solution.

The possibilities for the solution set for any homogeneous system is either a unique solution or infinitely many solutions. Since the homogeneous system has the zero solution and $x_1=3, x_2=-2, x_3=1$ is another solution, it has at least two distinct solution. Thus the only possibility is infinitely many solutions.

(h) A homogeneous system of $5$ equations in $3$ unknowns and the rank of the system is $3$.

A homogeneous system is always consistent. Since the rank $r$ of the system and the number $n$ of unknowns are equal, the only possibility is the zero solution (and the zero solution is a unique solution).

(i) A system of $3$ equations in $2$ unknowns and the rank of the system is $2$.

We don’t know whether the system is consistent or not.
If it is consistent, then since the rank $r$ and the number of unknowns are the same, the system has a unique solution. Thus the possibilities are either inconsistent or a unique solution.

Before talking about the rank, we need to discuss whether the system is inconsistent or not. For example, consider the following $3\times 2$ system
\begin{align*}
x_1+x_2&=1\\
2x_1+2x_2&=3\\
3x_1+3x_2&=3.
\end{align*}
Then the augmented matrix is
$\left[\begin{array}{rr|r} 1 & 1 & 1 \\ 2 &2 &3 \\ 3 & 3 & 3 \end{array}\right].$ We reduce this by elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rr|r}
1 & 1 & 1 \\
2 &2 &3 \\
3 & 3 & 3
\end{array}\right] \xrightarrow{\substack{R_2-2R_1\\R_3-3R_1}}
\left[\begin{array}{rr|r}
1 & 1 & 1 \\
0 & 0 &1 \\
0 & 0 &0
\end{array}\right].
\end{align*}
The last matrix is in echelon form and it has two nonzero rows. Thus, the rank of the system is $2$. However, the second row means that we have $0=1$. Hence the system is inconsistent.

(j) A homogeneous system of $4$ equations in $3$ unknowns and the rank of the system is $2$.

A homogeneous system is consistent. The rank is $r=2$ and the number of variables is $n=3$. Hence there is $n-r=1$ free variable. Thus there are infinitely many solutions.

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1. 02/13/2017

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