3 Systems of Linear Equations

From equations to matrices

A linear equation in variables $x_1,\ldots,x_n$ has the form

Several such equations impose conditions simultaneously. A solution is an ordered tuple $(x_1,\ldots,x_n)$ that satisfies every equation.

For example, the system

has coefficient matrix

unknown vector $\mathbf{x}$, and constants vector $\mathbf b$. Its compact form is $A\mathbf{x}=\mathbf b$. The corresponding augmented matrix is

The vertical bar separates coefficients from right-hand-side constants.

Takeaway: matrix notation organizes many equations into a form that can be simplified systematically.

Simplifying a system safely

The solution set is preserved by the three elementary row operations:

1. Interchange two rows.

2. Multiply a row by a nonzero constant.

3. Replace one row by itself plus a multiple of another row.

For example,

The matrix operation corresponds to replacing the second equation by the second equation minus three times the first. Because each operation is reversible, the original and transformed systems have exactly the same solutions.

Two matrices are row equivalent when one can be obtained from the other through a finite sequence of elementary row operations.

Takeaway: row reduction changes the appearance of a system, not its solution set.

Pivots and echelon forms

Row reduction aims to create a staircase pattern. In row-echelon form, zero rows appear below nonzero rows, each leading nonzero entry lies farther right than the leading entry above it, and entries below each leading entry are zero. Each leading nonzero entry is a pivot.

A pivot variable corresponds to a pivot column. A free variable corresponds to a nonpivot column. In reduced row-echelon form, every pivot equals $1$ and is the only nonzero entry in its column.

Gaussian elimination stops after reaching row-echelon form and then uses back-substitution. Gauss–Jordan elimination continues to reduced row-echelon form, where the solution can usually be read directly.

A practical procedure is:

1. Find the leftmost column containing a nonzero entry.

2. Move a nonzero entry into the current pivot position if necessary.

3. Scale the pivot row so the pivot is $1$, when convenient.

4. Use row replacement to create zeros below the pivot.

5. Continue in lower rows and columns to the right.

6. For reduced row-echelon form, eliminate entries above each pivot as well.

Takeaway: pivots reveal which variables are determined and which variables may remain free.

Reading and classifying solutions

The final matrix reveals the solution type. A row such as

represents $0=4$, which is impossible. Therefore, the system has no solution and is inconsistent.

If no contradictory row occurs, the system is consistent. Then:

• There is a unique solution when every variable column contains a pivot.

• There are infinitely many solutions when the system is consistent and at least one variable is free.

• There is no solution when row reduction produces $0=c$ with $c\neq 0$.

For example,

gives

The variable $z$ is free. Set $z=t$, giving

so the complete solution set is

Takeaway: count pivots and inspect the augmented column before deciding whether a system has one, infinitely many, or no solutions.

Consistency through rank

The rank of a matrix is the number of pivots in its row-reduced form. For an augmented system $A\mathbf{x}=\mathbf b$, consistency can be tested using

If the augmented matrix has a pivot in its final column, the augmented rank is larger and a contradictory row appears. If the ranks are equal, the system is consistent.

For a consistent system with $n$ variables:

• $\operatorname{rank}(A)=n$ means every variable column has a pivot, so the solution is unique.

• $\operatorname{rank}(A)<n$ means at least one variable is free, so infinitely many solutions result.

This rank viewpoint summarizes the pivot test and works for systems with different numbers of equations and variables.

Takeaway: rank compares the number of independent constraints with the number of unknowns.

A complete elimination example

Consider

Its augmented matrix is

Eliminating below the first pivot gives

Swap the second and third rows:

Then eliminate below the second pivot:

Back-substitution gives

and then

Thus,

There is a pivot in every variable column and no contradictory row, so the solution is unique.

Takeaway: record each row operation carefully, then interpret the final equations through back-substitution.

Applications and interpretation

Systems of linear equations model several quantities constrained at the same time.

In a mixture problem, let $x$ be the liters of a $20\%$ solution and $y$ the liters of a $50\%$ solution. To make $10$ liters containing $30\%$ acid, the conditions are

The first equation tracks total volume, and the second tracks total acid. Solving the system determines the amounts of each solution.

The same modeling pattern applies to network and flow problems, where conservation at a junction equates total incoming and outgoing flow. Economic models can use variables for quantities of goods and equations for budgets, capacities, or balance conditions.

A reliable modeling process is:

1. Define the unknown quantities.

2. Translate each condition into a linear equation.

3. Form the augmented matrix.

4. Row-reduce and interpret the result in context.

5. Reject mathematically valid values that violate practical restrictions, such as negative quantities or capacity limits.

Takeaway: the algebraic solution must be checked against the real-world meaning of the variables.

• system of linear equations