Solving 2 equations with 3 variables is a common scenario in linear algebra that appears whenever we have fewer independent equations than unknowns. Such systems are called underdetermined and typically have infinitely many solutions, which can be described using one or more free variables. Understanding how to find and express these solution sets is essential for fields ranging from engineering to economics, where models often contain more parameters than constraints.
Easier said than done, but still worth knowing.
Introduction
When we write a system of linear equations in the form
[ \begin{cases} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = b_1\[2pt] a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = b_2 \end{cases} ]
we have two equations but three unknowns ((x_1, x_2, x_3)). Consider this: unless the equations are contradictory, the solution set will not be a single point; instead it will be a line, a plane, or a higher‑dimensional affine subspace within (\mathbb{R}^3). The goal of solving 2 equations with 3 variables is to characterize that set in a clear, parametric form.
This changes depending on context. Keep that in mind Easy to understand, harder to ignore..
Why the System Is Underdetermined
The rank of the coefficient matrix tells us how many independent directions are constrained. With two rows, the maximum rank is 2. If the rank equals 2, there is one degree of freedom left (3 − 2 = 1 free variable). If the rank is smaller than 2 (e.g., the rows are multiples of each other), we have two or more free variables, leading to a plane or the whole space as the solution set. If the augmented matrix has a higher rank than the coefficient matrix, the system is inconsistent and has no solution.
Methods for Solving
Two standard techniques work well for these systems:
- Gaussian Elimination (Row Reduction) – transforms the augmented matrix to row‑echelon form, making the free variables obvious.
- Parameterization – after elimination, solve for the leading variables in terms of the free ones, then write the solution as a vector plus a scalar multiple of a direction vector.
Both approaches yield the same result; the choice depends on personal preference and the complexity of the coefficients.
Step‑by‑Step Gaussian Elimination
- Write the augmented matrix ([A|b]).
- Use elementary row operations to obtain a row‑echelon form:
- Swap rows if needed.
- Multiply a row by a non‑zero scalar.
- Add a multiple of one row to another.
- Identify pivot columns (leading variables) and non‑pivot columns (free variables).
- Express each leading variable as a linear combination of the free variables plus any constant term from the right‑hand side.
- Write the general solution in parametric vector form.
Parameterization Example
Suppose after elimination we obtain
[ \begin{aligned} x_1 &= 2 - 3t\ x_2 &= 5 + t\ x_3 &= t \end{aligned} ]
where (t) is a free parameter. The solution set can be written as
[ \begin{pmatrix}x_1\x_2\x_3\end{pmatrix}
\begin{pmatrix}2\5\0\end{pmatrix}
- t\begin{pmatrix}-3\1\1\end{pmatrix}, \qquad t\in\mathbb{R}. ]
Here (\begin{pmatrix}2\5\0\end{pmatrix}) is a particular solution and (\begin{pmatrix}-3\1\1\end{pmatrix}) spans the direction of the line of solutions.
Worked Examples
Example 1 – Homogeneous System
Solve
[ \begin{cases} x + 2y - z = 0\ 2x - y + 3z = 0 \end{cases} ]
Step 1 – Augmented matrix
[ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0\ 2 & -1 & 3 & 0 \end{array}\right] ]
Step 2 – Row operations
Replace (R_2) with (R_2 - 2R_1):
[ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0\ 0 & -5 & 5 & 0 \end{array}\right] ]
Divide (R_2) by (-5):
[ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0\ 0 & 1 & -1 & 0 \end{array}\right] ]
Replace (R_1) with (R_1 - 2R_2):
[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0\ 0 & 1 & -1 & 0 \end{array}\right] ]
Step 3 – Identify free variable
Column 3 has no pivot → (z) is free. Let (z = t).
Step 4 – Solve for leading variables
[ \begin{aligned} x + z &= 0 ;\Rightarrow; x = -t\ y - z &= 0 ;\Rightarrow; y = t \end{aligned} ]
Step 5 – Parametric form
[ \begin{pmatrix}x\y\z\end{pmatrix}
t\begin{pmatrix}-1\1\1\end{pmatrix}, \qquad t\in\mathbb{R}. ]
The solution set is a line through the origin spanned by ((-1,1,1)^T).
Example 2 – Non‑Homogeneous System
Solve
[ \begin{cases} 3x - y + 2z = 7\ x + 4y - z = -2 \end{cases} ]
Step 1 – Augmented matrix
[ \left[\begin{array}{ccc|c} 3 & -1 & 2 & 7\ 1 & 4 & -1 & -2 \end{array}\right] ]
Step 2 – Row operations
Swap rows to have a 1 in the top‑left:
[ \left[\begin{array}{ccc|c} 1 & 4 & -1 & -2\ 3 & -1 & 2 & 7 \end{array}\right] ]
Replace (R_2) with (R_2 - 3R_1):
[ \left[\begin{array}{ccc|c} 1 & 4 & -1 & -2\ 0 & -13 & 5 & 13 \end{array}\right] ]
Divide (R_2) by (-13):
[ \left[\begin{array}{ccc|c} 1 & 4 & -1 & -2\ 0 & 1 & -\frac{5}{13} & -1 \end{array}\right] ]
Replace (R_1) with (R_
