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Vector Equations and Matrix Equations

Systems of linear equations can be expressed as vector equations or as matrix equations. This different perspective allow us to prove that a system of linear equations cannot have exactly two solution.

Vector Equation and Matrix Equation

Consider a system of linear equations represented as an augmented matrix [Ab⃗][\begin{array}{c|c} A & \vec{b} \end{array}]. Suppose that the columns of AA are a⃗1,a⃗2,…,a⃗n\vec{a}_1,\vec{a}_2, \ldots,\vec{a}_n. Then, we can also represent it as

  1. A vector equation:
x1a⃗1+x2a⃗2+⋯+xna⃗n=b⃗x_1\vec{a}_1+x_2\vec{a}_2+\cdots+x_n\vec{a}_n = \vec{b}
  1. A matrix equation:
Ax⃗=b⃗A\vec{x} = \vec{b}

Consider the system of linear equations:

{2x−y+3z=8x+2y−z=1\begin{cases} 2x - y + 3z = 8 \\ x + 2y - z = 1 \end{cases}

We can rewrite this system as an equivalent vector equation:

x[21]+y[−12]+z[3−1]=[81]x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} -1 \\ 2 \end{bmatrix} + z\begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 8 \\ 1 \end{bmatrix}

Indeed, equality of vectors means equality in each coordinate. The equality of the first coordinates recovers the first linear equation, and the equality of the second coordinates recovers the second linear equation. Hence, this vector equation is equivalent to the original system.

A linear combination of vectors can be expressed as a matrix–vector product. Therefore, we can further rewrite the vector equation as the following matrix equation:

[2−1312−1][xyz]=[81]\begin{bmatrix}2 & -1 & 3\\ 1 & 2 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 8 \\ 1 \end{bmatrix}

The following forms represent the same system of equations.

1. System of linear equations: {x+2y−z=43x+y+2z=12x−y+z=0\begin{cases} x + 2y - z = 4 \\ 3x + y + 2z = 1 \\ 2x - y + z = 0 \end{cases}

2. Vector equation: x[132]+y[21−1]+z[−121]=[410]x\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} + z\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix}

3. Matrix equation: [12−13122−11][xyz]=[410]\begin{bmatrix} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix}

Why System of Linear Equations Cannot Have Exactly 2 Solutions

Using the matrix equation representation and the linearity of the matrix–vector product, we can establish a fundamental fact about linear systems:

Theorem:
If a system of linear equations has two distinct solutions, then it has infinitely many solutions.

Consequently, a system of linear equations can have zero, one, or infinitely many solutions, but it cannot have exactly two, three, or any other finite number greater than one.

Proof

We write the system of linear equations as a matrix equation Ax⃗=b⃗A\vec{x} = \vec{b}. Suppose x⃗1\vec{x}_1 and x⃗2\vec{x}_2 are two distinct solutions to Ax⃗=b⃗A\vec{x} = \vec{b}. We'll show that infinitely many solutions exist.

Step 1: Since both x⃗1\vec{x}_1 and x⃗2\vec{x}_2 are solutions:

Ax⃗1=b⃗andAx⃗2=b⃗A\vec{x}_1 = \vec{b} \quad \text{and} \quad A\vec{x}_2 = \vec{b}

Step 2: For any scalar tt, consider:

x⃗(t)=(1−t)x⃗1+tx⃗2\vec{x}(t) = (1-t)\vec{x}_1 + t\vec{x}_2

Step 3: Show that x⃗(t)\vec{x}(t) is also a solution using linearity:

Ax⃗(t)=A((1−t)x⃗1+tx⃗2)=A((1−t)x⃗1)+A(tx⃗2)(linearity)=(1−t)Ax⃗1+tAx⃗2(linearity)=(1−t)b⃗+tb⃗(both x⃗1,x⃗2 are solutions)=b⃗\begin{align*} A\vec{x}(t) &= A((1-t)\vec{x}_1 + t\vec{x}_2) \\ &= A((1-t)\vec{x}_1) + A(t\vec{x}_2) & \text{(linearity)} \\ &= (1-t)A\vec{x}_1 + tA\vec{x}_2 & \text{(linearity)} \\ &= (1-t)\vec{b} + t\vec{b} & \text{(both } \vec{x}_1, \vec{x}_2 \text{ are solutions)} \\ &= \vec{b} \end{align*}

Step 4: Since tt can be any real number, there are infinitely many solutions x⃗(t)\vec{x}(t).

The parametric form x⃗(t)=(1−t)x⃗1+tx⃗2\vec{x}(t) = (1-t)\vec{x}_1 + t\vec{x}_2 is exactly the line through x⃗1\vec{x}_1 and x⃗2\vec{x}_2:

  • Rewrite as: x⃗(t)=x⃗1+t(x⃗2−x⃗1)\vec{x}(t) = \vec{x}_1 + t(\vec{x}_2 - \vec{x}_1)
  • This is a line through x⃗1\vec{x}_1 in direction (x⃗2−x⃗1)(\vec{x}_2 - \vec{x}_1)

Key insight: The solution set to a system of linear equation is either:

  • Empty (no solutions)
  • A single point (one solution)
  • A line, plane, or hyperplane (infinitely many solutions)

It's never a finite set with more than one element. If it contains two distinct points, it also contains the line passing through those two points.

Consider the augmented matrix

[11331−111]\left[\begin{array}{ccc|c}1 & 1 & 3 & 3\\ 1 & -1 & 1 & 1\end{array}\right]

Suppose that we are given two distinct solutions

x⃗1=[210],x⃗2=[001]\vec{x}_1 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \quad \vec{x}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

Then, we can find more solutions by considering

x⃗(t)=(1−t)[210]+t[001]=[2(1−t)1−tt]=[2−2t1−tt]\vec{x}(t) = (1-t)\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2(1-t) \\ 1-t \\ t \end{bmatrix} = \begin{bmatrix} 2-2t \\ 1-t \\ t \end{bmatrix}

You can check yourself that the following vectors are solutions to the augmented matrix:

x⃗(2)=[−2−12];x⃗(−1)=[42−1]\vec{x}(2) = \begin{bmatrix} -2 \\ -1 \\ 2 \end{bmatrix}; \quad \vec{x}(-1) = \begin{bmatrix} 4 \\ 2 \\ -1 \end{bmatrix}