IB Mathematics AA – Topic 1: Number & Algebra

General Form of a 3×3 System:

where \(a_i, b_i, c_i, d_i\) are constants and \(x, y, z\) are the unknowns

Augmented Matrix Form:

The system above becomes:

The vertical line separates coefficients from constants

Each row represents one equation; each column (before the line) represents one variable

Three Main Approaches:

• Substitution/Elimination: Traditional algebra method (tedious for 3×3)
• Row operations (Gaussian elimination): Systematic matrix method (IB focus)
• Calculator/technology: Using GDC functions (fastest for IB exams)

In IB exams, you're expected to know row operations conceptually and use your calculator efficiently.

💡 Calculator Tips for IB:

• Learn the rref (reduced row echelon form) function on your GDC
• Enter the augmented matrix correctly—double-check dimensions
• The rref function automatically performs all row operations
• Final matrix shows solutions directly (or indicates no/infinite solutions)

Three Elementary Row Operations:

• 1. Row Swap (Interchange): \(R_i \leftrightarrow R_j\)

  Exchange two rows

  Example: Swap row 1 and row 2

• 2. Row Scaling (Multiplication): \(R_i \to kR_i\) (where \(k \neq 0\))

  Multiply all entries in a row by a nonzero constant

  Example: \(R_1 \to \frac{1}{2}R_1\) divides row 1 by 2

• 3. Row Addition (Replacement): \(R_i \to R_i + kR_j\)

  Add a multiple of one row to another row

  Example: \(R_2 \to R_2 - 3R_1\) subtracts 3 times row 1 from row 2

A matrix is in Row Echelon Form if:

• All rows consisting entirely of zeros are at the bottom
• The first nonzero entry (called a pivot) in each row is to the right of the pivot in the row above
• All entries below a pivot are zero

Example of REF:

A matrix is in Reduced Row Echelon Form if:

• It is in row echelon form
• Each pivot equals 1
• Each pivot is the only nonzero entry in its column

Example of RREF (unique solution):

⚠ Common Pitfalls:

• Incorrect row operations: Always replace the target row, not the source row
• Dividing by zero: Can't scale a row by zero; can't create a pivot from zero
• Losing track: Write down each operation you perform
• Matrix entry errors: Double-check when entering into calculator
• Misreading RREF: Know how to interpret the final form

Example 1: Solving Using Row Operations (Step-by-Step)

Problem: Solve the system using row operations:

Solution:

Step 1: Write as augmented matrix

\(\left[\begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 2 & 1 & -1 & 0 \\ 3 & -1 & 2 & 3 \end{array}\right]\)

Step 2: Eliminate \(x\) from rows 2 and 3

\(R_2 \to R_2 - 2R_1\):

New \(R_2\): \((2-2(1), 1-2(2), -1-2(1), 0-2(7)) = (0, -3, -3, -14)\)

\(R_3 \to R_3 - 3R_1\):

New \(R_3\): \((3-3(1), -1-3(2), 2-3(1), 3-3(7)) = (0, -7, -1, -18)\)

\(\left[\begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 0 & -3 & -3 & -14 \\ 0 & -7 & -1 & -18 \end{array}\right]\)

Step 3: Scale row 2 to get pivot of 1

\(R_2 \to -\frac{1}{3}R_2\):

\(\left[\begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 0 & 1 & 1 & \frac{14}{3} \\ 0 & -7 & -1 & -18 \end{array}\right]\)

Step 4: Eliminate \(y\) from row 3

\(R_3 \to R_3 + 7R_2\):

\(\left[\begin{array}{ccc|c} 1 & 2 & 1 & 7 \\ 0 & 1 & 1 & \frac{14}{3} \\ 0 & 0 & 6 & \frac{44}{3} \end{array}\right]\)

Step 5: Scale row 3 and back-substitute

\(R_3 \to \frac{1}{6}R_3\): third row becomes \([0, 0, 1, \frac{22}{9}]\)

Continue to RREF (eliminate upward)...

Final RREF: \(\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & \frac{22}{9} \end{array}\right]\)

Solution: \(x = 1\), \(y = 2\), \(z = \frac{22}{9}\)

Note: In IB exams, use your calculator's rref function to skip these tedious steps!

Three Solution Types

Quick Recognition Guide:

1. Count the pivots (leading 1's) in RREF:
   • 3 pivots in 3 columns → unique solution
   • 2 or fewer pivots → either no solution or infinite solutions
2. Check for contradictions (rows like \([0 \; 0 \; 0 \;|\; k]\) with \(k \neq 0\)):
   • If present → no solution
   • If absent → infinite solutions (when fewer than 3 pivots)

⚠ Common Exam Mistakes:

• Missing infinitely many solutions: A row of zeros (without contradiction) means free variables!
• Wrong interpretation: \([0 \; 0 \; 0 \;|\; 0]\) is NOT a contradiction—it's consistent
• Not expressing parametric form: For infinite solutions, write \(x, y, z\) in terms of parameter \(t\)
• Calculation errors: Double-check matrix entries in calculator

Example 2: System with No Solution

Problem: Determine the number of solutions and find them if they exist:

Solution:

Step 1: Write augmented matrix

\(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 4 & -2 & 8 \\ 1 & 2 & -1 & 5 \end{array}\right]\)

Step 2: Apply row operations

\(R_2 \to R_2 - 2R_1\):

\(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 0 & 0 & 2 \\ 1 & 2 & -1 & 5 \end{array}\right]\)

Step 3: Identify contradiction

Row 2 reads: \(0x + 0y + 0z = 2\)

This simplifies to: \(0 = 2\) ← CONTRADICTION!

Conclusion: NO SOLUTION (inconsistent system)

Geometric interpretation: The three planes do not have a common intersection point. Equations 1 and 3 represent parallel planes (same normal vector, different constants).

Example 3: System with Infinitely Many Solutions (IB-Style)

Problem: Solve the system and express the solution in parametric form:

Solution:

Step 1: Write augmented matrix

\(\left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 2 & 4 & 6 & 12 \\ 1 & 2 & 3 & 6 \end{array}\right]\)

Step 2: Apply row operations to RREF

Notice: Row 2 = 2 × Row 1, and Row 3 = Row 1

After row operations:

\(\left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\)

Step 3: Identify free variables

Only one pivot (in first column). This means \(y\) and \(z\) are free variables.

Let \(y = s\) and \(z = t\) (parameters)

Step 4: Express solution in parametric form

From row 1: \(x + 2y + 3z = 6\)

Substitute parameters: \(x + 2s + 3t = 6\)

Solve for \(x\): \(x = 6 - 2s - 3t\)

Solution (parametric form):
\(x = 6 - 2s - 3t\)
\(y = s\)
\(z = t\)
where \(s, t \in \mathbb{R}\)

Alternative vector form:
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} + s\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix}\)

Geometric interpretation: The three equations represent the same plane (or parallel coincident planes), so any point on this plane is a solution—infinitely many!

Using Your Calculator:

• Learn the rref function location on your GDC before the exam
• Enter matrices carefully: check dimensions match (3×4 for augmented 3×3)
• Interpret the output correctly: look for contradictions or free variables
• Write down the RREF matrix in your answer for full marks

Common Exam Questions:

• "Solve the system": Find unique solution or state no solution/infinite solutions
• "Determine the number of solutions": Identify type (1, 0, or ∞)
• "Express in parametric form": Use parameters for free variables
• "Find the value of k...": Modify system to change solution type

Time Management:

• Systems questions typically worth 4-6 marks
• Using calculator efficiently can save 3-5 minutes per question
• Always verify your answer makes sense (substitute back if time permits)