IB Mathematics AA Systems of Liner Equations Study Notes
IB Mathematics AA Systems of Liner Equations Study Notes
IB Mathematics AA Systems of Liner Equations Study Notes Offer a clear explanation of Systems of Liner Equations , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Systems of Liner Equations.
Systems of Linear Equations
The study of systems of linear equations is fundamental to understanding mathematical relationships and solving real-world problems. This topic focuses on solving systems of linear equations, including unique solutions, infinite solutions, and inconsistent systems. The syllabus encourages the use of both algebraic and technological methods, such as row reduction or matrix operations.
Key Concepts:
- Definition of a Linear Equation:
A linear equation is an equation of the first degree, meaning the highest power of the variable(s) is 1.
Example: ax + by + cz = d, where a, b, c, d are constants, and x, y, z are variables.
- System of Linear Equations:
A set of two or more linear equations involving the same variables.
Example:
- x + y + z = 6
- 2x – y + 3z = 14
- -x + 2y – z = -2
Methods of Solving Systems of Linear Equations:
- Algebraic Methods:
- Substitution Method: Solve one equation for one variable and substitute it into the other equations.
- Elimination Method: Eliminate one variable by adding or subtracting equations, reducing the system to fewer equations.
- Gaussian Elimination: Use row operations to reduce the system to row-echelon form, then solve using back-substitution.
- Matrix Methods:
- Represent the system as a matrix AX = B, where A is the coefficient matrix, X is the variable column matrix, and B is the constant matrix.
- Solve using row reduction, inverse matrix, or computational tools.
- Technological Methods: Use graphing calculators or software (e.g., MATLAB, Python, or Wolfram Alpha) for larger or complex systems.
Nature of Solutions:
- Unique Solution: The system has one solution, meaning all equations intersect at a single point.
- Infinite Solutions: The system has infinitely many solutions, typically represented by parametric equations.
- No Solution (Inconsistent System): The system is contradictory, such as parallel lines or planes that do not intersect.
Applications:
- Intersection of Lines and Planes: Understanding the intersection points or regions in three-dimensional geometry.
- Real-World Applications: Optimizing resources in engineering, solving equilibrium problems in physics, and managing economics models.
Example : Unique Solution
Given the system of equations:
- x + y + z = 6
- 2x – y + 3z = 14
- -x + 2y – z = -2
Step-by-step Solution:
Write the equations:
\(\quad x + y + z = 6 \\ \quad 2x – y + 3z = 14 \\ \quad -x + 2y – z = -2 \)
Eliminate \(x\) using equations (2) and (3):
From equation (1), multiply and subtract from equation (2):
\(3y – z = -2 \quad \text{(4)} \)
From equation (1), add to equation (3):
\(y – 2z = -8 \quad \text{(5)} \)
Solve the two-variable system (4) and (5):
Combine equations to find \(y\):
\(y = 2 \quad \text{(6)}\)
Substitute \(y = 2\) into equation (4):
\(z = 8 \quad \text{(7)} \)
Find \(x\):
Substitute \(y = 2\) and \(z = 8\) into equation (1):
\(x = -4 \quad \text{(8)} \)
Final Solution: x = -4, y = 2, z = 8
IB Mathematics AA SL Systems of Linear Equations Exam Style Worked Out Questions
[Maximum mark: 6]
Question:
Consider any three consecutive integers, n – 1, n and n + 1.
(a) Prove that the sum of these three integers is always divisible by 3.
▶️Answer/Explanation
Ans: (n – 1) + n + (n + 1)
= 3n
which is always divisible by 3
(b) Prove that the sum of the squares of these three integers is never divisible by 3.
▶️Answer/Explanation
Ans: (n -1)2 +n2 + (n+1)2 (=n2 – 2n + 1 + n2 + n2 + 2n + 1)
attempts to expand either (n -1)2 or (n+1)2 ( do not accept n2 – 1 or n2 + 1)
= 3n2 + 2
demonstrating recognition that 2 is not divisible by 3 or \(\frac{2}{3}\) seen after correct
expression divided by 3
3n2 is divisible by 3 and so 3n2 + 2 is never divisible by 3
OR the first term is divisible by 3, the second is not
OR \(3\left ( n^{2}+\frac{2}{3} \right ) OR \frac{3n^{2} + 2}{3} = n^{2} + \frac{2}{3}\)
hence the sum of the squares is never divisible by 3
Question
Consider the following equations, where a , \(b \in \mathbb{R}:\)
\(x + 3y + (a – 1)z = 1\)
\(2x + 2y + (a – 2)z = 1\)
\(3x + y + (a – 3)z = b.\)
a. If each of these equations defines a plane, show that, for any value of a , the planes do not intersect at a unique point.[3]
▶️Answer/Explanation
Ans:
METHOD 1
\(\det \left( {\begin{array}{*{20}{c}}
1&3&{a – 1} \\
2&2&{a – 2} \\
3&1&{a – 3}
\end{array}} \right)\) M1
\( = 1\left( {2(a – 3) – (a – 2)} \right) – 3\left( {2(a – 3) – 3(a – 2)} \right) + (a – 1)(2 – 6)\)
(or equivalent) A1
= 0 (therefore there is no unique solution) A1
[3 marks]
METHOD 2
\(\left( {\begin{array}{*{20}{c}}
1&3&{a – 1}\\
2&2&{a – 2}\\
3&1&{a – 3}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
1\\
b. Find the value of b for which the intersection of the planes is a straight line.[4]
▶️Answer/Explanatio
Ans:
\end{array}} \right):\left( {\begin{array}{*{20}{c}}
1&3&{a – 1}\\
0&{ – 4}&{ – a}\\
0&{ – 8}&{ – 2a}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{b – 3}
\end{array}} \right)\) M1A1
\(:\left( {\begin{array}{*{20}{c}}
1&3&{a – 1}\\
0&{ – 4}&{ – a}\\
0&0&0
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{b – 1}
\end{array}} \right)\) (and 3 zeros imply no unique solution) A1
[3 marks]
METHOD 1
\(\left( {\begin{array}{*{20}{c}}
1&3&{a – 1}\\
2&2&{a – 2}\\
3&1&{a – 3}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
1\\
b
\end{array}} \right):\left( {\begin{array}{*{20}{c}}
1&3&{a – 1}\\
0&{ – 4}&{ – a}\\
0&{ – 8}&{ – 2a}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{b – 3}
\end{array}} \right)\) M1A1
\(:\left( {\begin{array}{*{20}{c}}
1&3&{a – 1}\\
0&{ – 4}&{ – a}\\
0&0&0
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{b – 1}
\end{array}} \right)\) A1
b = 1 A1 N2
Note: Award M1 for an attempt to use row operations.
[4 marks]
METHOD 2
b = 1 A4
Note: Award A4 only if “ b −1 ” seen in (a).
[4 marks]