# Possibilities For the Number of Solutions for a Linear System

## Problem 102

Determine whether the following systems of equations (or matrix equations) described below has no solution, one unique solution or infinitely many solutions and justify your answer.

**(a)** \[\left\{

\begin{array}{c}

ax+by=c \\

dx+ey=f,

\end{array}

\right.

\]
where $a,b,c, d$ are scalars satisfying $a/d=b/e=c/f$.

**(b)** $A \mathbf{x}=\mathbf{0}$, where $A$ is a singular matrix.

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

**(d) **$A\mathbf{x}=\mathbf{b}$, where the row-reduced echelon form of the augmented matrix $[A|\mathbf{b}]$ looks as follows:

\[\begin{bmatrix}

1 & 0 & -1 & 0 \\

0 &1 & 2 & 0 \\

0 & 0 & 0 & 1

\end{bmatrix}.\]
(*The Ohio State University, Linear Algebra Exam*)

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## Hint.

Recall that possibilities for the solution set of a system of linear equation is either no solution (inconsistent) or one unique solution or infinitely many solutions.

A homogeneous system is a system with zero constant terms.

A homogeneous system always has the zero solution.

## Solution.

**(a)** Note that by the given condition $a/d=b/e=c/f$, the numbers $b, e, c$ are not zero. The augmented matrix of the system is

$\left[\begin{array}{rr|r}

a & b & c \\

d & e & f

\end{array}\right]$.

We apply the elementary row operations as follows.

\begin{align*}

&\left[\begin{array}{rr|r}

a & b & c \\

d & e & f

\end{array}\right]
\xrightarrow{R_1-\frac{a}{d}R_2}

\left[\begin{array}{rr|r}

0 & 0 & 0 \\

d & e & f

\end{array}\right]
\xrightarrow{R_1\leftrightarrow R_2}

\left[\begin{array}{rr|r}

d & e & f\\

0 & 0 & 0

\end{array}\right]\\

& \xrightarrow{\frac{1}{d}}

\left[\begin{array}{rr|r}

1 & e/d & f/d\\

0 & 0 & 0

\end{array}\right].

\end{align*}

Here in the first step, we used the relation $a/d=b/e=c/f$.

The last matrix is in reduced row echelon form with rank $1$ and it does not have a row of the form $[00|1]$. Therefore the system is consistent and there are $1(=\text{the number of unknowns }-\text{ rank})$ free variable, hence there are infinitely many solutions.

** (b)** The system is homogeneous, thus it has the zero solution. Since the coefficient matrix $A$ is singular, the system has non-zero solution as well.

Therefore, the only possibility is that the system has infinitely many solutions.

** (c)** A homogeneous system has the zero solution hence it is consistent. Since there are more unknowns than equations, the system must have infinitely many solutions.

** (d)** The last row of the reduced row echelon form matrix of the augmented matrix is $[000|1]$.

It corresponds to the equation

\[0x_1+0x_2+0x_3=1.\]
Equivalently, this is $0=1$ and this is impossible. Thus the system has no solution (an inconsistent system).

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