Cheatsheet 06 - Systems of Linear Equations
Mathematics for Master Students
What “Linear” Means
An equation is linear when every unknown appears on its own, to the first power: a weighted sum of the variables equals a number. No powers, no products of variables, no roots or exponents on a variable.
| Linear | Not linear |
|---|---|
| \(2x + 3y = 13\) | \(x^2 + y = 4\) (a square) |
| \(x_1 + 2x_2 + 2x_3 = 16\) | \(xy = 6\) (a product) |
| \(4a + 7b - c = 0\) | \(\sqrt{x} + y = 1\) (a root) |
A system asks several linear equations to hold at the same time. A solution is a set of values satisfying every equation at once. Always check a candidate against all of them.
Solving a 2×2 System
Substitution. Solve one equation for one variable, drop it into the other:
- Isolate a variable, e.g. \(x = 5 - y\).
- Substitute into the other equation and solve for the remaining variable.
- Back-substitute to get the first variable.
Elimination. Scale the equations so a variable cancels:
- Multiply one (or both) equations so a variable has matching coefficients.
- Add or subtract to cancel that variable; solve for what remains.
- Back-substitute for the other variable.
Both methods give the same point where the two lines cross. Always verify in the original equations.
The Augmented Matrix
A system is just its numbers. Stack the coefficients, then the right-hand side after a bar. For example, \(\;x + y + z = 6,\ \ x + 2y + 3z = 11,\ \ 2x + y + z = 9\;\) becomes
\[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 11 \\ 2 & 1 & 1 & 9 \end{array}\right]\]
Each row is an equation; the bar separates left from right. No matrix algebra: pure bookkeeping.
Each one keeps the solution set unchanged:
- Swap two rows (reorder the equations).
- Scale a row by a non-zero number (multiply or divide an equation).
- Add a multiple of one row to another (combine two equations).
Every operation acts on the whole row (the right-hand side included).
Gaussian Elimination Recipe
To solve a \(3 \times 3\) system:
- Forward-eliminate: using row operations, create zeros below the diagonal, one column at a time, until the matrix is triangular (a staircase of zeros in the lower-left).
- Read the triangular system from the bottom up: the last row gives one variable directly.
- Back-substitute upward, one variable at a time.
- Verify the triple in the original equations.
Micro-example (from the matrix above): \[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 2 \end{array}\right] \;\Rightarrow\; z = 2,\ \ y + 2(2) = 5 \Rightarrow y = 1,\ \ x + 1 + 2 = 6 \Rightarrow x = 3.\]
How Many Solutions?
The triangular form tells you the fate without a picture:
| In the matrix you see | Number of solutions | Geometry (2 unknowns) |
|---|---|---|
| A leading entry in every column | exactly one | lines cross at a point |
| A row \(\;0 = (\text{non-zero})\), e.g. \(0 = 5\) | none | parallel, distinct lines |
| A row \(\;0 = 0\) (whole row vanishes) | infinitely many | the same line |
\(\;0 = 0\;\) means an equation was redundant → infinitely many solutions. “No solution” appears only as \(\;0 = (\text{non-zero})\), a true contradiction. Never confuse the two.
Parameterize the free case: when a row vanishes, set the free variable to \(t\). For \(x - 3y = 2\) with a redundant partner: let \(y = t\), then \(x = 2 + 3t\), giving the solution set \((x, y) = (2 + 3t,\; t)\).
Market Equilibrium
A market clears where supply equals demand, one price and quantity solving both curves:
\[\text{demand } q = 90 - 3p, \quad \text{supply } q = 2p - 10 \;\Longrightarrow\; 90 - 3p = 2p - 10 \;\Rightarrow\; p = 20,\ q = 30.\]
Set the two expressions for \(q\) equal (or eliminate \(q\) from the system), solve for \(p\), then back-substitute for \(q\). Below equilibrium there is excess demand (shortage); above it, excess supply (surplus). The same idea drives production-planning systems.
Systems Checklist
- Is every equation actually linear? (first power, no products or roots)
- Did you apply each row operation to the whole row, right-hand side included?
- Did you reduce to triangular form before back-substituting?
- Did you read \(\;0 = 0\;\) (infinitely many) vs \(\;0 = (\text{non-zero})\) (none) correctly?
- Did you verify the solution in the original equations?
This is the last cheatsheet. Bring all six to the Mock Exam.