📘 Chapter 4: Matrix Based System of Linear Equations | NEB Mathematics Notes Class 12
Consistency, independence, dependence & classification of linear systems
🔢 Matrix Based System of Linear Equations covers solving and classifying systems of linear equations. A system can be consistent (has at least one solution) or inconsistent (no solution). If consistent, it can be independent (unique solution) or dependent (infinitely many solutions). Below are detailed classifications with step-by-step reasoning.
📖 1. Classification Rules
For two linear equations:
\(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\)
• Consistent & Independent (Unique solution): \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
• Consistent & Dependent (Infinite solutions): \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
• Inconsistent (No solution): \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
\(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\)
• Consistent & Independent (Unique solution): \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
• Consistent & Dependent (Infinite solutions): \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
• Inconsistent (No solution): \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
🎯 2. Classify the following systems
1a \(4x – 3y = -6\) \(-4x + 2y = 16\)
Step 1: Write equations in standard form:
\(4x – 3y = -6\) … (i)
\(-4x + 2y = 16\) … (ii)
Step 2: Compare coefficients: \(a_1 = 4, b_1 = -3, c_1 = -6\)
\(a_2 = -4, b_2 = 2, c_2 = 16\)
Step 3: Check ratios: \(\frac{a_1}{a_2} = \frac{4}{-4} = -1\), \(\frac{b_1}{b_2} = \frac{-3}{2} = -1.5\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the system has a unique solution.
\(4x – 3y = -6\) … (i)
\(-4x + 2y = 16\) … (ii)
Step 2: Compare coefficients: \(a_1 = 4, b_1 = -3, c_1 = -6\)
\(a_2 = -4, b_2 = 2, c_2 = 16\)
Step 3: Check ratios: \(\frac{a_1}{a_2} = \frac{4}{-4} = -1\), \(\frac{b_1}{b_2} = \frac{-3}{2} = -1.5\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the system has a unique solution.
✅ Consistent and Independent
1b \(9x – 2y = -4\) \(3x + 4y = 1\)
Step 1: \(a_1 = 9, b_1 = -2, c_1 = -4\)
\(a_2 = 3, b_2 = 4, c_2 = 1\)
Step 2: \(\frac{a_1}{a_2} = \frac{9}{3} = 3\), \(\frac{b_1}{b_2} = \frac{-2}{4} = -0.5\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), unique solution exists.
\(a_2 = 3, b_2 = 4, c_2 = 1\)
Step 2: \(\frac{a_1}{a_2} = \frac{9}{3} = 3\), \(\frac{b_1}{b_2} = \frac{-2}{4} = -0.5\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), unique solution exists.
✅ Consistent and Independent
1c \(-6x + 4y = 10\) \(3x – 2y = -5\)
Step 1: Multiply second equation by -2: \(3x – 2y = -5\) becomes \(-6x + 4y = 10\)
Step 2: Both equations become identical: \(-6x + 4y = 10\)
\(a_1 = -6, b_1 = 4, c_1 = 10\) and \(a_2 = -6, b_2 = 4, c_2 = 10\)
Step 3: \(\frac{a_1}{a_2} = 1\), \(\frac{b_1}{b_2} = 1\), \(\frac{c_1}{c_2} = 1\)
All ratios are equal, so infinite solutions.
Step 2: Both equations become identical: \(-6x + 4y = 10\)
\(a_1 = -6, b_1 = 4, c_1 = 10\) and \(a_2 = -6, b_2 = 4, c_2 = 10\)
Step 3: \(\frac{a_1}{a_2} = 1\), \(\frac{b_1}{b_2} = 1\), \(\frac{c_1}{c_2} = 1\)
All ratios are equal, so infinite solutions.
✅ Consistent and Dependent
1d \(3x – 4y = 1\) \(6x – 8y = 7\)
Step 1: \(a_1 = 3, b_1 = -4, c_1 = 1\)
\(a_2 = 6, b_2 = -8, c_2 = 7\)
Step 2: \(\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{-4}{-8} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{1}{7}\)
Step 3: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Therefore, no solution exists.
\(a_2 = 6, b_2 = -8, c_2 = 7\)
Step 2: \(\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{-4}{-8} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{1}{7}\)
Step 3: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Therefore, no solution exists.
✅ Inconsistent (No solution)
1e \(25x – 15y = 45\) \(-5x – 3y = 24\)
Step 1: Simplify first equation: divide by 5 → \(5x – 3y = 9\)
Second equation: \(-5x – 3y = 24\)
Step 2: \(a_1 = 5, b_1 = -3, c_1 = 9\)
\(a_2 = -5, b_2 = -3, c_2 = 24\)
Step 3: \(\frac{a_1}{a_2} = \frac{5}{-5} = -1\), \(\frac{b_1}{b_2} = \frac{-3}{-3} = 1\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), unique solution.
Second equation: \(-5x – 3y = 24\)
Step 2: \(a_1 = 5, b_1 = -3, c_1 = 9\)
\(a_2 = -5, b_2 = -3, c_2 = 24\)
Step 3: \(\frac{a_1}{a_2} = \frac{5}{-5} = -1\), \(\frac{b_1}{b_2} = \frac{-3}{-3} = 1\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), unique solution.
✅ Consistent and Independent
📊 3. Additional Classifications
2a \(2x + 3y = 5\) \(4x + 6y = 10\)
\(\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2}\)
All ratios equal → infinite solutions.
All ratios equal → infinite solutions.
✅ Consistent and Dependent
2b \(x + 2y = 3\) \(2x + 4y = 7\)
\(\frac{a_1}{a_2} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{3}{7}\)
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → No solution.
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → No solution.
✅ Inconsistent (No solution)
2c \(3x – y = 4\) \(6x – 2y = 8\)
\(\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{4}{8} = \frac{1}{2}\)
All ratios equal → infinite solutions (second equation is 2 times first).
All ratios equal → infinite solutions (second equation is 2 times first).
✅ Consistent and Dependent
2d \(5x – 2y = 3\) \(10x – 4y = 5\)
\(\frac{a_1}{a_2} = \frac{5}{10} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{3}{5}\)
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → No solution.
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → No solution.
✅ Inconsistent (No solution)
2e \(x + y = 5\) \(x – y = 1\)
\(\frac{a_1}{a_2} = \frac{1}{1} = 1\), \(\frac{b_1}{b_2} = \frac{1}{-1} = -1\)
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) → Unique solution (x=3, y=2).
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) → Unique solution (x=3, y=2).
✅ Consistent and Independent
🔧 4. Solving Linear Systems (Matrix Method)
3 Solve: \(x + y = 5\), \(x – y = 1\)
Method 1 (Elimination): Add equations: \(2x = 6\) ⇒ \(x = 3\)
Substitute: \(3 + y = 5\) ⇒ \(y = 2\)
Matrix form: \(\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}\)
Determinant = \(-1 – 1 = -2 \neq 0\) → Unique solution.
Substitute: \(3 + y = 5\) ⇒ \(y = 2\)
Matrix form: \(\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}\)
Determinant = \(-1 – 1 = -2 \neq 0\) → Unique solution.
✅ \(x = 3, y = 2\)
4 Solve: \(2x + 3y = 8\), \(4x + 6y = 16\)
Second equation is 2 times first: \(4x + 6y = 16\) ⇒ \(2(2x + 3y) = 2(8)\)
Both equations represent the same line. General solution: \(2x + 3y = 8\) ⇒ \(y = \frac{8 – 2x}{3}\)
Infinite solutions. Let \(x = t\), then \(y = \frac{8 – 2t}{3}\)
Both equations represent the same line. General solution: \(2x + 3y = 8\) ⇒ \(y = \frac{8 – 2x}{3}\)
Infinite solutions. Let \(x = t\), then \(y = \frac{8 – 2t}{3}\)
✅ Infinitely many solutions: \(x = t, y = \frac{8-2t}{3}, t \in \mathbb{R}\)
5 Solve: \(x + 2y = 3\), \(2x + 4y = 7\)
Multiply first equation by 2: \(2x + 4y = 6\)
Second equation: \(2x + 4y = 7\)
Both left sides are equal but right sides differ (6 ≠ 7).
∴ No solution.
Second equation: \(2x + 4y = 7\)
Both left sides are equal but right sides differ (6 ≠ 7).
∴ No solution.
✅ No solution (Inconsistent system)
📝 5. Quick Reference Table
| Condition | Nature | Number of Solutions |
|---|---|---|
| \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) | Consistent & Independent | Unique solution |
| \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) | Consistent & Dependent | Infinite solutions |
| \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) | Inconsistent | No solution |
🧠 Matrix Based System of Linear Equations – Key Points
📌 A system is consistent if it has at least one solution.
📌 A system is inconsistent if it has no solution (parallel lines).
📌 For infinite solutions, the equations are multiples of each other.
📌 For unique solution, lines intersect at exactly one point.
📌 Matrices and determinants provide systematic methods for solving systems (Cramer’s Rule, Inverse Matrix).