🧮 Solving Linear Equations

Solving a system of linear equations means isolating each variable without changing the system’s meaning

A system of linear equations is simplified by moving the focus away from the variables of the system of equations by organising the coefficients by their variables into a matrix

Coefficients and constants are separated by a line in this matrix, known as the augmented matrix

To simplify and solve the system, we apply Elementary Row Operations and reduce the variables systematically until all but one of each variable is left

🧭 Directionality and Convention

Because we write left to right, top to bottom, matrix reduction follows the same ergonomic flow:

Start at the top-left
Move rightward and downward
At each step, keep the current coefficient representing the variable and eliminate the rest above and below it

The Pivot The coefficient of the variable that we isolate for each row is called the pivot

The term “pivot” comes from its role in elimination:

It’s the anchor used to eliminate entries below (and in RREF, above) it.
In Gaussian elimination, we “pivot” around this entry to clear its column.
It acts like a hinge or fulcrum—the transformation rotates around it.

This naturally leads to a staggered structure that looks like an echelon arranged by row, thus it is called row-echelon form, or REF for short

What is “Echelon”? The term echelon comes from military formation—soldiers arranged in a staggered diagonal pattern.

🔄 Reduction Pipeline

Reducing both directions at once is rather disorienting, thus we have preferred to go in one direction preferred at a time

Since we start top-left, we naturally eliminate downwards first (forward-substitution), to remove all the variables below the leading variable of each row

Row-Echelon Form

$$ \left[ \begin{array}{ccc|c} 2 & 4 & 6 & 8 \\ 0 & 3 & 9 & 6 \\ 0 & 0 & 5 & 15 \end{array} \right] $$

All nonzero rows are above any rows of all zeros.
Each pivot is to the right of the pivot in the row above.
All entries below a pivot are zeros.

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Elimination Checkpoint Elimination that stops here is called Gaussian Elimination

Once we hit the bottom we eliminate upwards (backward-substitution), to remove the all the variables above the leading variable of each row

Then afterwards, we make the coefficients of each isolated variable into 1, and we have reached the solution to the system of linear equations

Reduced Row Echelon Form At this point, the diagonal of isolated coefficients of 1 with the rest of the matrix being 0, is called the Reduced Row-Echelon Form, basically Reduced REF, or just RREF
$$ \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -7 \\ 0 & 0 & 1 & 3 \end{array} \right] $$
Each pivot is 1
Each pivot is the only nonzero entry in its column
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Elimination Endpoint Elimination that reaches here is called Jordan-Gauss Elimination

🧠 General Strategy for Solving to REF or RREF

Here’s a robust tactic for reducing any matrix:

🔁 Swap rows to move already solved rows (with pivot = 1) to the top.
🔽 Work top-down, row by row.
🧹 Eliminate entries below the pivot using row operations.
🔁 Repeat for the next row and pivot column.
🧼 For RREF: After reaching REF, eliminate entries above each pivot and normalize pivots to 1.

Normalize each pivot to 1 immediately before using it to eliminate other rows
This leads to cleaner arithmetic and aligns naturally with RREF goals.
Favor additive operations over scalar multiplication
Instead of computing $cR_1$, use $R_2 + cR_1$ or $R_1 + cR_2$ to eliminate—this keeps the original row intact and avoids unnecessary scaling.

✏️ Example: Normalizing First for Cleaner Elimination

Start with:

$$ \begin{bmatrix} 3 & -4 & 10 \\ -5 & 8 & -17 \\ -3 & 12 & -12 \end{bmatrix} $$

Step-by-step:

Normalize $R_1$:

$$ R_1 \rightarrow \frac{1}{3}R_1 $$

Eliminate below using clean multiples:

$$ R_2 + 5R_1,\quad R_3 + 3R_1 $$

Normalize $R_2$:

$$ R_2 \rightarrow \frac{3}{4}R_2 $$

Eliminate below again:

$$ R_3 - 8R_2 $$

Final REF:

$$ \begin{bmatrix} 1 & -\frac{4}{3} & \frac{10}{3} \\ 0 & 1 & -\frac{1}{4} \\ 0 & 0 & 0 \end{bmatrix} $$

This method avoids fractional multipliers like $\frac{5}{3}R_1$ and keeps operations visually clean and semantically aligned with pivot logic.

🧠 Interpreting the Final Augmented Matrix: 3 Possibilities

After reducing the augmented matrix (via REF or RREF), there are three possible outcomes:

✅ 1. Unique Solution

Every variable has a pivot.
No contradictions.
Final matrix has one solution.

$$ \begin{bmatrix} 1 & 0 & 0 & \vert & 2 \\ 0 & 1 & 0 & \vert & -1 \\ 0 & 0 & 1 & \vert & 3 \end{bmatrix} $$

→ Solution: $x = 2$, $y = -1$, $z = 3$

♾️ 2. Infinitely Many Solutions

Some variables are free (no pivot in their column).
No contradictions.
Final matrix leads to parametric form.

$$ \begin{bmatrix} 1 & 0 & 2 & \vert & 5 \\ 0 & 1 & -1 & \vert & 3 \\ 0 & 0 & 0 & \vert & 0 \end{bmatrix} $$

→ Let $z = t$ (free variable) Then: $x = 5 - 2t$ $y = 3 + t$

→ Parametric solution: $(x, y, z) = (5 - 2t,\ 3 + t,\ t)$

❌ 3. No Solution

A row like:

$$ [0 \quad 0 \quad 0 \mid c] \quad \text{where } c \neq 0 $$

This represents a contradiction (e.g., $0 = 5$).
System is inconsistent.

$$ \begin{bmatrix} 1 & 2 & -1 & \vert & 4 \\ 0 & 1 & 3 & \vert & 2 \\ 0 & 0 & 0 & \vert & 7 \end{bmatrix} $$

→ Contradiction in last row: $0 = 7$ → ❌ No solution

Last updated on 2025-11-08

🧮 Parameterisation 🧱 Matrix Partitioning