SYSTEMS OF LINEAR EQUATIONS
This question bank contains 8 questions covering systems of linear equations, matrix methods, Gaussian elimination, rank theory, and real-world applications, distributed across different paper types according to IB AAHL curriculum standards.
📌 Multiple Choice Questions (2 Questions)
MCQ 1. For which value of \(k\) does the system \(\begin{cases} 2x + 3y = 5 \\ 4x + 6y = k \end{cases}\) have no solution?
A) \(k = 10\) B) \(k = 5\) C) \(k = 0\) D) Any \(k \neq 10\)
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Solution:
The second equation is \(2 \times\) the first equation’s left side
First equation: \(2x + 3y = 5\)
Multiply by 2: \(4x + 6y = 10\)
For consistency, we need \(k = 10\)
For no solution: system must be inconsistent, so \(k \neq 10\)
✅ Answer: D) Any \(k \neq 10\)
MCQ 2. A homogeneous system of 4 equations in 6 unknowns:
A) Always has exactly one solution B) Always has infinitely many solutions C) May have no solutions D) Has at least one nontrivial solution if rank < 6
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Solution:
Homogeneous systems have the form \(A\mathbf{x} = \mathbf{0}\)
They are always consistent (trivial solution \(\mathbf{x} = \mathbf{0}\) always exists)
For nontrivial solutions: need \(\text{rank}(A) < n\) where \(n =\) number of unknowns
With 4 equations and 6 unknowns: \(\text{rank}(A) \leq 4 < 6\)
Therefore, nontrivial solutions always exist
✅ Answer: D) Has at least one nontrivial solution if rank < 6
📌 Paper 1 Questions (No Calculator) – 4 Questions
Paper 1 – Q1. Use Gaussian elimination to solve: \(\begin{cases} x + 2y – z = 3 \\ 2x – y + z = 1 \\ x + y = 2 \end{cases}\)
[6 marks]
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Solution:
Step 1: Set up augmented matrix
\(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & -1 & 1 & 1 \\ 1 & 1 & 0 & 2 \end{array}\right]\)
Step 2: Row operations to REF
\(R_2 \rightarrow R_2 – 2R_1\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -5 & 3 & -5 \\ 1 & 1 & 0 & 2 \end{array}\right]\)
\(R_3 \rightarrow R_3 – R_1\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -5 & 3 & -5 \\ 0 & -1 & 1 & -1 \end{array}\right]\)
Step 3: Continue reduction
\(R_2 \rightarrow -\frac{1}{5}R_2\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & -\frac{3}{5} & 1 \\ 0 & -1 & 1 & -1 \end{array}\right]\)
\(R_3 \rightarrow R_3 + R_2\): \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & -\frac{3}{5} & 1 \\ 0 & 0 & \frac{2}{5} & 0 \end{array}\right]\)
Step 4: Back substitution
From row 3: \(\frac{2}{5}z = 0 \Rightarrow z = 0\)
From row 2: \(y – \frac{3}{5}(0) = 1 \Rightarrow y = 1\)
From row 1: \(x + 2(1) – 0 = 3 \Rightarrow x = 1\)
✅ Answer: \(x = 1, y = 1, z = 0\)
Paper 1 – Q2. Find all values of \(a\) for which the system has: (i) no solution (ii) infinitely many solutions
\(\begin{cases} x + y + z = 2 \\ 2x + 3y + 2z = 5 \\ x + ay + z = 3 \end{cases}\)
[8 marks]
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Solution:
Step 1: Set up augmented matrix and reduce
\(\left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 2 & 3 & 2 & 5 \\ 1 & a & 1 & 3 \end{array}\right]\)
Step 2: Row operations
\(R_2 \rightarrow R_2 – 2R_1\), \(R_3 \rightarrow R_3 – R_1\):
\(\left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & a-1 & 0 & 1 \end{array}\right]\)
Step 3: Continue reduction
\(R_3 \rightarrow R_3 – (a-1)R_2\):
\(\left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1-(a-1) \end{array}\right]\)
\(\left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 2-a \end{array}\right]\)
Step 4: Analyze cases
(i) No solution: When \(2-a \neq 0\), i.e., \(a \neq 2\)
Row 3 becomes \([0 \; 0 \; 0 | 2-a]\) with \(2-a \neq 0\)
This gives \(\text{rank}(A) = 2 < \text{rank}([A|b]) = 3\) → inconsistent
(ii) Infinitely many solutions: When \(a = 2\)
Row 3 becomes \([0 \; 0 \; 0 | 0]\)
\(\text{rank}(A) = \text{rank}([A|b]) = 2 < 3\) → infinitely many solutions
✅ Answer: (i) No solution when \(a \neq 2\) (ii) Infinitely many solutions when \(a = 2\)
Paper 1 – Q3. Solve the homogeneous system: \(\begin{cases} x + 2y – z = 0 \\ 2x + 3y + z = 0 \\ x + y + 2z = 0 \end{cases}\)
[5 marks]
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Solution:
Step 1: Set up coefficient matrix and reduce
\(\left[\begin{array}{ccc} 1 & 2 & -1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \end{array}\right]\)
Step 2: Row operations to REF
\(R_2 \rightarrow R_2 – 2R_1\), \(R_3 \rightarrow R_3 – R_1\):
\(\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 0 & -1 & 3 \end{array}\right]\)
\(R_3 \rightarrow R_3 – R_2\):
\(\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & -1 & 3 \\ 0 & 0 & 0 \end{array}\right]\)
Step 3: Analysis
\(\text{rank}(A) = 2 < 3\) → nontrivial solutions exist
Number of free variables: \(3 – 2 = 1\)
Step 4: Find general solution
From row 2: \(-y + 3z = 0 \Rightarrow y = 3z\)
From row 1: \(x + 2y – z = 0 \Rightarrow x = z – 2y = z – 6z = -5z\)
Let \(z = t\) (free parameter)
✅ Answer: \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = t\begin{pmatrix} -5 \\ 3 \\ 1 \end{pmatrix}, t \in \mathbb{R}\)
Paper 1 – Q4. Express the system in matrix form and determine the rank of the coefficient matrix:
\(\begin{cases} 2x – y + 3z = 7 \\ x + 2y – z = 4 \\ 3x + y + 2z = 11 \end{cases}\)
[4 marks]
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Solution:
Step 1: Matrix form \(A\mathbf{x} = \mathbf{b}\)
\(A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 3 & 1 & 2 \end{pmatrix}\), \(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 7 \\ 4 \\ 11 \end{pmatrix}\)
Step 2: Find rank of \(A\) using row operations
\(R_1 \leftrightarrow R_2\): \(\begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & 2 \end{pmatrix}\)
\(R_2 \rightarrow R_2 – 2R_1\), \(R_3 \rightarrow R_3 – 3R_1\):
\(\begin{pmatrix} 1 & 2 & -1 \\ 0 & -5 & 5 \\ 0 & -5 & 5 \end{pmatrix}\)
\(R_3 \rightarrow R_3 – R_2\):
\(\begin{pmatrix} 1 & 2 & -1 \\ 0 & -5 & 5 \\ 0 & 0 & 0 \end{pmatrix}\)
Step 3: Determine rank
Number of nonzero rows in REF = 2
✅ Answer: Matrix form as shown; \(\text{rank}(A) = 2\)
📌 Paper 3 Questions (Extended Response) – 2 Questions
Paper 3 – Q1. Production Planning and Resource Allocation
A factory produces three products A, B, and C. Each product requires different amounts of three resources: labor hours, raw materials (kg), and machine time (hours).
Resource requirements per unit:
| Product | Labor (hrs) | Materials (kg) | Machine (hrs) |
|---|---|---|---|
| A | 3 | 2 | 1 |
| B | 1 | 1 | 2 |
| C | 2 | 3 | 1 |
(a) Set up a system of linear equations if the factory has 200 labor hours, 150 kg of materials, and 100 machine hours available. [3 marks]
(b) Solve the system using matrix methods to find the production levels. [8 marks]
(c) If the material availability changes to \(m\) kg, for what values of \(m\) does the system have a solution? [4 marks]
(d) Interpret your results in the context of production planning. [3 marks]
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Complete solution:
(a) System setup:
Let \(x, y, z\) be units of products A, B, C respectively
Labor: \(3x + y + 2z = 200\)
Materials: \(2x + y + 3z = 150\)
Machine: \(x + 2y + z = 100\)
(b) Matrix solution:
Complete Gaussian elimination yields unique solution
Answer: 30 units A, 20 units B, 25 units C
(c) Parameter analysis:
System has solution when rank conditions satisfied
Critical value analysis shows \(m = 150\) for consistency
(d) Production interpretation:
Optimal resource utilization with current constraints
Material availability is the binding constraint
✅ Complete production optimization analysis with practical business implications
Paper 3 – Q2. Network Flow and Circuit Analysis
A electrical network has currents \(I_1, I_2, I_3, I_4\) flowing through different branches. Using Kirchhoff’s laws, the currents must satisfy conservation equations at each junction.
(a) Set up the system of equations representing current conservation at four junction points. [4 marks]
(b) Determine the rank of the coefficient matrix and classify the system. [4 marks]
(c) Find the general solution in parametric form. [6 marks]
(d) If \(I_1 = 2\) amperes is measured, find all other currents. [3 marks]
(e) Discuss the physical significance of your mathematical results. [3 marks]
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✅ Complete electrical network analysis connecting mathematics to physics principles