True or False Quiz About a System of Linear Equations
Problem 78
Determine whether the following sentence is True or False.
(Purdue University Linear Algebra Exam)
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Contents
- Which of the following statements are true?
- Solution.
- (a) True or False: A linear system of four equations in three unknowns is always inconsistent.
- (b) True or False: A linear system with fewer equations than unknowns must have infinitely many solutions.
- (c) True or False:If the system $A\mathbf{x}=\mathbf{b}$ has a unique solution, then $A$ must be a square matrix.
Which of the following statements are true?
(a) A linear system of four equations in three unknowns is always inconsistent.
(b) A linear system with fewer equations than unknowns must have infinitely many solutions.
(c) If the system $A\mathbf{x}=\mathbf{b}$ has a unique solution, then $A$ must be a square matrix.
Solution.
All of them are false as we explain below.
(a) True or False: A linear system of four equations in three unknowns is always inconsistent.
Consider any homogeneous system of four linear equations and three unknowns. Since a homogeneous system always has the solution $\mathbf{x}=\mathbf{0}$. Thus the statement (a) is false.
As an explicit example, the homogeneous system
\[\left\{
\begin{array}{c}
x+y+z=0 \\
2x+2y+2z=0 \\
3x+3y+3z=0
\end{array}
\right.
\]
has the solution $(x,y,z)=(0,0,0)$. So the system is consistent.
(b) True or False: A linear system with fewer equations than unknowns must have infinitely many solutions.
Consider the system of one equation with two unknowns
\[0x+0y=1.\]
This system has no solution at all. Hence the statement is false.
(c) True or False:If the system $A\mathbf{x}=\mathbf{b}$ has a unique solution, then $A$ must be a square matrix.
Consider the matrix $A=\begin{bmatrix}
1 \\
1
\end{bmatrix}$. Then the system
\[\begin{bmatrix}
1 \\
1
\end{bmatrix}[x]=\begin{bmatrix}
0 \\
0
\end{bmatrix}\]
has the unique solution $x=0$ but $A$ is not a square matrix.
A more theoretical argument is as follows. If vectors $\mathbf{v}_1,\dots, \mathbf{v}_k$ are linearly independent, then the system
\[[\mathbf{v}_1 \dots \mathbf{v}_l]\mathbf{x}=\mathbf{0}\]
has the unique solution
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