The document discusses elementary row operations and row echelon form (REF) and reduced row echelon form (RREF) for matrices. It describes the row reduction algorithm to convert a matrix into REF and RREF forms using elementary row operations. The algorithm proceeds by finding the leftmost nonzero pivot column at each step and using row operations to create zeros below the pivot. Converting a matrix to RREF form allows determining whether a linear system has no solution, a unique solution, or infinitely many solutions.

1. Math 337 Lecture #1

2. Elementary Row Operations
• Row Replacement:
– row = row + multiple of another row
• Row Interchange
– Switch two rows
• Scaling
– Multiply all entries of a row by a constant (≠0)
• Row operations are reversible
• Matrices row equivalent  same soln set

3. Row Echelon Form (REF)
• All nonzero rows above any rows of all zeros
• Each leading entry (left-most non-zero entry –
PIVOT position) of a row is in a column to the
right of the leading entry above it
• All entries in a column below a leading entry
are zeros
• REF is NOT unique

4. Reduced Row Echelon Form (RREF)
• Leading entry in each non-zero row is 1
• Each leading entry is only non-zero entry in its
column
• RREF is unique

5. Row Reduction Algorithm
• Step 1: Left most nonzero column is a pivot
column. Pivot position is at top of column
• Step 2: Select a nonzero entry in the pivot column
as a pivot. If necessary, interchange rows to move
this entry into the pivot position
• Step 3: Use row replacement to create zeros in all
positions below the pivot.
• Step 4: Cover (ignore) the row containing the
pivot & any rows above it. Appy steps 1-3 to
submatrix that remains.

6. Row Reduction Algorithm - Backward
• To go to RREF:
• Step 5: Begin with rightmost pivot & work
backward (to the left) and upward, using row
replacement to create zeros in above each
pivot. If a pivot is not 1, make it 1 by a scaling
operation

7. Solutions to Linear Systems
• No solution  inconsistent
• Single solution  consistent
• Many solutions  consistent
– General solution  basic vars in terms of free vars

8. Existence & Uniqueness Thereom
• A linear system is consistent iff the rightmost
column of the augmented matrix is not a pivot
column, i.e., iff the REF form of the
augmented matrix has no rows of the form:
– [0 0 0 … 0 b]

9. Consistent Systems
• Either:
• Unique solution  no free variables
• Infinitely many solutions  free variables exist

10. Process for Solving Systems of Equns
1. Write augmented matrix
2. Reduce to REF
– Consistent?  consistent?
3. Reduce to RREF
4. Write system corresponding to 3
5. Solve equations for basic vars in terms of free
variables.
no
Yes – keep going