For Each Final Matrix State The Solution

For each final matrix state the solution is the key idea that links the outcome of Gaussian elimination (or Gauss‑Jordan elimination) to the actual answer of a linear system. When you finish row‑reducing an augmented matrix, the shape of that final matrix tells you whether the system has a single solution, infinitely many solutions, or none at all. Understanding how to read that final state is essential for anyone studying linear algebra, engineering, computer science, or any field that relies on solving sets of equations. The following article walks you through the theory, the step‑by‑step procedure, concrete examples, and common questions so you can confidently state the solution for every possible final matrix configuration.

Introduction

The phrase for each final matrix state the solution captures a fundamental skill in linear algebra: after performing row operations until the matrix is in a canonical form (usually row echelon form or reduced row echelon form), you must interpret that form to write down the solution set of the original system. This interpretation depends on three possibilities—unique solution, infinite solutions, or no solution—and each corresponds to a distinct pattern in the final matrix. By mastering the patterns and the reasoning behind them, you turn a mechanical elimination process into a clear, logical answer.

Understanding Matrix Forms

Before we state the solution, we need to recognize what the final matrix looks like. Two forms are most useful:

Row Echelon Form (REF)

A matrix is in row echelon form when:

• All nonzero rows are above any rows of all zeros.
• The leading entry (the first nonzero number from the left, also called a pivot) of each nonzero row is to the right of the pivot of the row above it.
• All entries in a column below a pivot are zero.

Reduced Row Echelon Form (RREF)

A matrix is in reduced row echelon form when it satisfies the REF conditions plus:

• Each pivot is equal to 1.
• Each pivot is the only nonzero entry in its column.

Most textbooks ask you to go all the way to RREF because it makes reading the solution immediate, but even REF is enough to decide consistency and to parameterize free variables.

Steps to Determine the Solution from a Final Matrix

Regardless of whether you stop at REF or continue to RREF, the following four steps let you state the solution for each final matrix state.

Step 1: Identify Pivot Positions

Scan the matrix from left to right, top to bottom. Every leading 1 (or leading nonzero entry in REF) marks a pivot. The column containing a pivot corresponds to a basic variable; columns without pivots correspond to free variables.

Step 2: Check for Inconsistencies

Look for any row that looks like

[0 0 … 0 | b]   with b ≠ 0```

In other words, a row where all coefficients are zero but the augmented entry is non‑zero. Such a row encodes the equation \(0 = b\), which is impossible. If you find one, the system is **inconsistent** and has **no solution**.

### Step 3: Determine Free Variables  

If the system is consistent, count the number of pivots (the rank, \(r\)). Suppose the original system has \(n\) variables. Then the number of free variables is \(n - r\). Assign parameters (commonly \(t_1, t_2, \dots\)) to each free variable.

### Step 4: Write the Solution Set  

Express each basic variable in terms of the free variables using the equations represented by the nonzero rows. If there are no free variables, the solution is a single vector (unique solution). If there are one or more free variables, the solution set is an affine subspace: a particular solution plus a linear combination of vectors that span the null space.

These steps work for **each final matrix state** because the state (REF or RREF) uniquely reveals the pivot pattern, any contradictory rows, and the relationship between basic and free variables.

## Scientific Explanation: Why the Final Matrix Determines the Solution  

The power of the final matrix comes from two core theorems in linear algebra.

### Rank and Nullity  

For an \(m \times n\) coefficient matrix \(A\), the **rank** (\(\text{rank}(A)\)) equals the number of pivots in its REF. The **nullity** (\(\dim(\text{Nul}(A))\)) equals \(n - \text{rank}(A)\). The Rank‑Nullity Theorem tells us that the solution space of the homogeneous system \(A\mathbf{x}= \mathbf{0}\) has dimension equal to the nullity. When we add an augmented column to form \([A|\mathbf{b}]\), the same rank tells us whether \(\mathbf{b}\) lies in the column space of \(A\). If \(\text{rank}([A|\mathbf{b}]) = \text{rank}(A)\), the system is consistent; otherwise, it is inconsistent.

### Consistency Theorem  

A linear system \(A\mathbf{x} = \mathbf{b}\) is consistent **iff** the augmented matrix does not contain a row of the form \([0\ \dots\ 0 \mid b]\) with \(b \neq 0\). This is exactly what Step 2 checks. When consistency holds, the set of solutions is an affine translate of \(\text{Nul}(A)\): \(\mathbf{x} = \mathbf{x}_p + \text{Span}\{\mathbf{v}_1,\dots,\mathbf{v}_{n-r}\}\), where \(\mathbf{x}_p\) is any particular solution obtained by back‑substitution.

Thus, the final matrix encodes both the rank (through pivots) and the augmented column’s relationship to the coefficient columns (through possible contradictory rows). Reading those features directly yields the solution set.

## Examples  

To solidify the procedure, consider three typical final matrices (shown in RREF for clarity) and state the solution for each.

### Example 1: Unique Solution  

\[
\left[\begin{array}{ccc|c}
1 & 0 & 0 & 4\\
0 & 1 & 0 & -2\\
0 & 0 & 1 & 7
\end{array}\right]
\]

- **Pivots**: columns 1, 2, 3 → rank = 3.  
- **No zero

rows in the augmented part → consistent.

- **Free variables**: None.

- **Solution**: \(\mathbf{x} = \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix}\). This is the unique solution.

### Example 2: Infinite Solutions with One Free Variable

\[
\left[\begin{array}{ccc|c}
1 & 0 & 2 & 1\\
0 & 1 & 3 & 2\\
0 & 0 & 0 & 0
\end{array}\right]
\]

- **Pivots**: columns 1, 2 → rank = 2.

- **Augmented matrix**: consistent (no \([0\ \dots\ 0 \mid b]\) with \(b \neq 0\)).

- **Free variables**: \(z\) (column 3).

- **Solution**: \(\mathbf{x} = \begin{pmatrix} 1 - 2z \\ 2 - 3z \\ z \end{pmatrix}\). Parametrize \(z\) to get the general solution.

### Example 3: Inconsistent System

\[
\left[\begin{array}{ccc|c}
1 & 2 & 0 & 3\\
0 & 0 & 1 & 4\\
0 & 0 & 0 & 1
\end{array}\right]
\]

- **Pivots**: columns 1, 3 → rank = 2.

- **Augmented matrix**: inconsistent (last row \([0\ 0\ 0\ |\ 1]\)).

- **Solution**: No solution exists.

## Conclusion

The process of solving linear systems using row reduction is both powerful and systematic. By transforming the augmented matrix into row-echelon form or reduced row-echelon form, we can identify the pivot positions, determine the rank, and check for consistency. This method provides a clear pathway to understanding whether a system has a unique solution, infinitely many solutions, or no solution at all. The final matrix state encapsulates all necessary information to derive the solution set, making it an indispensable tool in linear algebra. Whether dealing with theoretical problems or practical applications, mastering row reduction and interpreting the final matrix is crucial for solving linear systems efficiently and accurately.