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Linear Algebra Unit 1: Systems of linear equations

A linear system can have one solution, none, or infinitely many. Row reduction is the method that decides which and finds them.

Covers: Gaussian elimination, row echelon form, solution sets

Study guide

A linear system is a set of linear equations in the same variables. Write it as an augmented matrix, where each row is an equation and the last column holds the constants. Solutions don't change under the three elementary row operations: swapping two rows, multiplying a row by a nonzero constant, and adding a multiple of one row to another.

Gaussian elimination uses these operations to reach row echelon form: all-zero rows at the bottom, and each row's leading entry (pivot) to the right of the one above it. Continuing to reduced row echelon form makes each pivot a 1 with zeros above and below it. The RREF of a matrix is unique.

Reading the solution set: a row like [0 0 0 | 3] means 0 = 3, so the system is inconsistent with no solution. Otherwise the system is consistent. Columns with pivots correspond to basic variables, and columns without pivots to free variables. No free variables means exactly one solution; at least one free variable means infinitely many, written in parametric vector form with the free variables as parameters.

A linear system always has zero, one or infinitely many solutions, never exactly two. A homogeneous system Ax = 0 always has the trivial solution x = 0, and it has nontrivial solutions exactly when there's a free variable.

To find which right-hand sides b make Ax = b solvable, row reduce the augmented matrix [A | b] with b's entries left as symbols. Each zero row of A's echelon form gives a condition: the matching combination of b₁, b₂, … must equal 0. Every solution of a consistent system Ax = b has the form x = x_p + x_h, where x_p is any one particular solution and x_h ranges over all solutions of Ax = 0. So the solution set is the null space shifted by x_p.

When an entry depends on a parameter, row reduce as usual and watch for pivots that become 0 for particular parameter values. At those values, check the right-hand side: a nonzero entry beside a zero row means no solution, and a zero entry means a free variable and infinitely many solutions. For an m×n matrix of rank r, Ax = b has at most one solution for every b exactly when r = n (no free variables), and it has a solution for every b exactly when r = m (a pivot in every row). When r = m = n the solution always exists and is unique.

Key ideas

Formulas and rules

Pivot columns ↔ basic variables
Non-pivot columns ↔ free variables

Common mistakes

Words to know

Three elementary row operations
Swap two rows, multiply a row by a nonzero constant, add a multiple of one row to another.
Inconsistent row (0 = nonzero)
0 = 3: the system has no solution.
Free variable
A variable without a pivot; a consistent system then has infinitely many solutions.
Reduced row echelon form
Leading 1s in each pivot, with zeros above and below every pivot.
Number of solutions (linear systems)
Always zero, one or infinitely many.
Linear system
A set of linear equations in the same variables.
Augmented matrix
The coefficient matrix with the constants as a last column.
Row echelon form
Zero rows at the bottom; each pivot to the right of the one above.
Pivot
The leading nonzero entry in a row of an echelon form.
Consistent system
A system with at least one solution.
Homogeneous system
Ax = 0; always has the trivial solution.
Parametric vector form
The solution written as a vector plus free variables times vectors.
Gaussian elimination
Row reducing to echelon form to solve a system.

Sample practice questions

27 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

A row [0 0 0 | 4] appears after row reduction. The system has…

  1. Exactly one solution
  2. No solution
  3. x = 4
  4. Infinitely many solutions
Show the answer and steps

Answer: B) No solution

  1. The row says 0x + 0y + 0z = 4.
  2. 0 = 4 is impossible.
  3. No solution.
Question 2 (level 2)

Solve x + y = 3 and 2x − y = 0.

  1. (1, 2)
  2. (1, 1)
  3. (3, 0)
  4. (2, 1)
Show the answer and steps

Answer: A) (1, 2)

  1. Add the equations: 3x = 3, so x = 1.
  2. 1 + y = 3, so y = 2.
  3. (1, 2).
Question 3 (level 3)

A consistent system has a 3 × 4 coefficient matrix with 3 pivots. Number of free variables?

  1. 1
  2. 3
  3. 0
  4. 4
Show the answer and steps

Answer: A) 1

  1. 4 columns = 4 variables; 3 have pivots.
  2. Free variables = 4 − 3 = 1.