# 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).

For which choice(s) of the constant $k$ is the following matrix invertible? \[A=\begin{bmatrix} 1 & 1 & 1 \\ 1...