Multivariable Linear Systems
Extending our methods to three or more variables using systematic elimination.
Introduction
In 3D space, we need three equations to pin down a single point. Gaussian elimination systematically reduces a system to triangular form, where the last equation has only one variable. From there, weback-substitute upward to find all unknowns—a technique that scales to systems of any size.
Prerequisite Connection
You can solve 2×2 systems using substitution and elimination.
Today's Increment
We extend to 3 variables using Gaussian elimination and back-substitution.
Why This Matters
3D geometry, circuit analysis, and multivariable calculus all require systems with 3+ unknowns.
Key Concepts
Triangular (Row Echelon) Form
Each equation has one fewer variable than the one above.
Back-Substitution Steps
Solve the LAST equation for its variable (easiest: )
Substitute into the second-to-last equation:
Continue upward until all variables are found
Worked Examples
Example 1: Back-Substitution (Basic)
The system is already in triangular form:
Step 1: From equation 3
Step 2: Substitute into equation 2
Step 3: Substitute into equation 1
Solution:
Example 2: Full Gaussian Elimination (Intermediate)
Solve:
Step 1: Eliminate from equations 2 and 3
Eq2 - 2×Eq1:
Eq3 - Eq1:
Step 2: Eliminate
Step 3: Back-substitute
Solution:
Example 3: Dependent System (Advanced)
Solve:
Observation
Equation 2 is just 2× Equation 1 → only 2 independent equations for 3 unknowns
Result
Infinitely many solutions along a line in 3D space. Express solutions in terms of a parameter .
Parametric Solution: for any
Common Pitfalls
Arithmetic errors during elimination
With 3 variables, there are many operations—double-check each step carefully.
Wrong back-substitution order
Always start from the LAST equation (with fewest variables) and work upward.
Missing dependent systems
If you get , check for infinitely many solutions (parametric form).
Real-World Application
Electrical Circuit Analysis (Kirchhoff's Laws)
Engineers use systems of 3+ equations to find currents in complex circuits. Each loop and junction generates one equation, and solving the system gives the current through each component.
Example: A circuit with 3 loops produces 3 equations in 3 unknowns ().
Practice Quiz
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