📊 → [A|B]
📊 Chapter 16: System of Linear Equations | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Gaussian Elimination, Cramer’s Rule, Matrix Method
✅ Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
📊 System of Linear Equations — This chapter covers solving systems of linear equations using various methods including substitution, elimination, Gaussian elimination, Cramer’s Rule, and matrix methods. The notes have been updated according to the latest syllabus of 2080. Now you don’t need to go anywhere searching for the notes of this chapter because we are here to serve you.
📚 Exercises
📖 Introduction
A System of Linear Equations is a set of two or more linear equations involving the same set of variables. A linear equation is an equation of the first degree (highest power of any variable is 1).
General form: \[ a_1x + b_1y + c_1z = d_1 \] \[ a_2x + b_2y + c_2z = d_2 \] \[ a_3x + b_3y + c_3z = d_3 \]
General form: \[ a_1x + b_1y + c_1z = d_1 \] \[ a_2x + b_2y + c_2z = d_2 \] \[ a_3x + b_3y + c_3z = d_3 \]
🔧 Methods of Solving Linear Systems
1. Substitution Method: Solve one equation for one variable and substitute into others.
2. Elimination Method: Add/subtract equations to eliminate variables.
3. Gaussian Elimination: Transform to row echelon form.
4. Cramer’s Rule: Using determinants for unique solutions.
5. Matrix Method: \(AX = B\) ⇒ \(X = A^{-1}B\)
📐 Gaussian Elimination Method
Steps:
Row Operations:
- Write the augmented matrix [A|B]
- Use row operations to get upper triangular form
- Perform back substitution to find variables
Row Operations:
- Swap two rows
- Multiply a row by a non-zero constant
- Add a multiple of one row to another
📌 Consistency of Linear Systems
Consistent: Has at least one solution
• Unique solution (independent)
• Infinite solutions (dependent)
• Unique solution (independent)
• Infinite solutions (dependent)
Inconsistent: No solution (parallel lines/planes)
🧮 Cramer’s Rule
For system \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\):
\[
D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}, \quad
D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}, \quad
D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}
\]
\[
x = \frac{D_x}{D}, \quad y = \frac{D_y}{D} \quad (D \neq 0)
\]
📝 Solved Problems (15 Important Questions)
Q1 Solve by substitution: \(x + y = 5\), \(x – y = 1\).
From first: \(y = 5 – x\). Substitute into second: \(x – (5 – x) = 1\) ⇒ \(2x – 5 = 1\) ⇒ \(2x = 6\) ⇒ \(x = 3\)
Then \(y = 5 – 3 = 2\)
Then \(y = 5 – 3 = 2\)
✅ \(x = 3, y = 2\)
Q2 Solve by elimination: \(2x + 3y = 8\), \(3x – 2y = -1\).
Multiply first by 2: \(4x + 6y = 16\). Multiply second by 3: \(9x – 6y = -3\)
Add: \(13x = 13\) ⇒ \(x = 1\). Then \(2(1) + 3y = 8\) ⇒ \(3y = 6\) ⇒ \(y = 2\)
Add: \(13x = 13\) ⇒ \(x = 1\). Then \(2(1) + 3y = 8\) ⇒ \(3y = 6\) ⇒ \(y = 2\)
✅ \(x = 1, y = 2\)
Q3 Using Cramer’s Rule: \(3x + 4y = 10\), \(2x – y = 3\).
\(D = \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} = 3(-1) – 4(2) = -3 – 8 = -11\)
\(D_x = \begin{vmatrix} 10 & 4 \\ 3 & -1 \end{vmatrix} = 10(-1) – 4(3) = -10 – 12 = -22\) ⇒ \(x = \frac{-22}{-11} = 2\)
\(D_y = \begin{vmatrix} 3 & 10 \\ 2 & 3 \end{vmatrix} = 3(3) – 10(2) = 9 – 20 = -11\) ⇒ \(y = \frac{-11}{-11} = 1\)
\(D_x = \begin{vmatrix} 10 & 4 \\ 3 & -1 \end{vmatrix} = 10(-1) – 4(3) = -10 – 12 = -22\) ⇒ \(x = \frac{-22}{-11} = 2\)
\(D_y = \begin{vmatrix} 3 & 10 \\ 2 & 3 \end{vmatrix} = 3(3) – 10(2) = 9 – 20 = -11\) ⇒ \(y = \frac{-11}{-11} = 1\)
✅ \(x = 2, y = 1\)
Q4 Solve using matrix method: \(x + 2y = 5\), \(3x – y = 1\).
\(A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}\), \(B = \begin{bmatrix} 5 \\ 1 \end{bmatrix}\)
\(A^{-1} = \frac{1}{(1)(-1)-(2)(3)} \begin{bmatrix} -1 & -2 \\ -3 & 1 \end{bmatrix} = \frac{1}{-7} \begin{bmatrix} -1 & -2 \\ -3 & 1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}\)
\(X = A^{-1}B = \frac{1}{7} \begin{bmatrix} 1(5) + 2(1) \\ 3(5) + (-1)(1) \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 7 \\ 14 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\)
\(A^{-1} = \frac{1}{(1)(-1)-(2)(3)} \begin{bmatrix} -1 & -2 \\ -3 & 1 \end{bmatrix} = \frac{1}{-7} \begin{bmatrix} -1 & -2 \\ -3 & 1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}\)
\(X = A^{-1}B = \frac{1}{7} \begin{bmatrix} 1(5) + 2(1) \\ 3(5) + (-1)(1) \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 7 \\ 14 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\)
✅ \(x = 1, y = 2\)
Q5 Solve by Gaussian elimination: \(x + y + z = 6\), \(2x – y + z = 3\), \(x + 2y – z = 3\).
Augmented matrix: \(\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & -1 & 1 & | & 3 \\ 1 & 2 & -1 & | & 3 \end{bmatrix}\)
R2 → R2 – 2R1: \(\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -3 & -1 & | & -9 \\ 0 & 1 & -2 & | & -3 \end{bmatrix}\)
R3 → R3 – R1: \(\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -3 & -1 & | & -9 \\ 0 & 1 & -2 & | & -3 \end{bmatrix}\)
R3 → R3 + \(\frac{1}{3}\)R2: gives \(x=1, y=2, z=3\)
R2 → R2 – 2R1: \(\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -3 & -1 & | & -9 \\ 0 & 1 & -2 & | & -3 \end{bmatrix}\)
R3 → R3 – R1: \(\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -3 & -1 & | & -9 \\ 0 & 1 & -2 & | & -3 \end{bmatrix}\)
R3 → R3 + \(\frac{1}{3}\)R2: gives \(x=1, y=2, z=3\)
✅ \(x = 1, y = 2, z = 3\)
Q6 Check consistency: \(2x + 3y = 5\), \(4x + 6y = 10\).
Second equation = 2 × first equation ⇒ infinite solutions.
✅ Consistent and dependent (infinite solutions)
Q7 Check consistency: \(x + 2y = 3\), \(2x + 4y = 7\).
Multiply first by 2: \(2x + 4y = 6\). Second equation: \(2x + 4y = 7\) ⇒ contradiction.
✅ Inconsistent (no solution)
Q8 Find value of k for which system has unique solution: \(x + 2y = 3\), \(3x + ky = 5\).
For unique solution, \(\frac{1}{3} \neq \frac{2}{k}\) ⇒ \(k \neq 6\)
✅ \(k \neq 6\)
Q9 Find k for infinite solutions: \(2x + 3y = 5\), \(4x + ky = 10\).
For infinite solutions, \(\frac{2}{4} = \frac{3}{k} = \frac{5}{10}\) ⇒ \(\frac{1}{2} = \frac{3}{k}\) ⇒ \(k = 6\)
✅ \(k = 6\)
Q10 Solve system graphically: \(x + y = 4\), \(2x – y = 2\).
First: \(y = 4 – x\). Second: \(y = 2x – 2\). Intersection: \(4 – x = 2x – 2\) ⇒ \(6 = 3x\) ⇒ \(x = 2\), \(y = 2\)
✅ \(x = 2, y = 2\)
Q11 Using Cramer’s Rule: \(2x – 3y = 7\), \(5x + y = 9\).
\(D = \begin{vmatrix} 2 & -3 \\ 5 & 1 \end{vmatrix} = 2(1) – (-3)(5) = 2 + 15 = 17\)
\(D_x = \begin{vmatrix} 7 & -3 \\ 9 & 1 \end{vmatrix} = 7(1) – (-3)(9) = 7 + 27 = 34\) ⇒ \(x = 34/17 = 2\)
\(D_y = \begin{vmatrix} 2 & 7 \\ 5 & 9 \end{vmatrix} = 2(9) – 7(5) = 18 – 35 = -17\) ⇒ \(y = -17/17 = -1\)
\(D_x = \begin{vmatrix} 7 & -3 \\ 9 & 1 \end{vmatrix} = 7(1) – (-3)(9) = 7 + 27 = 34\) ⇒ \(x = 34/17 = 2\)
\(D_y = \begin{vmatrix} 2 & 7 \\ 5 & 9 \end{vmatrix} = 2(9) – 7(5) = 18 – 35 = -17\) ⇒ \(y = -17/17 = -1\)
✅ \(x = 2, y = -1\)
Q12 Solve: \(x + y + z = 6\), \(x – y + z = 2\), \(x + y – z = 0\).
Add first and second: \(2x + 2z = 8\) ⇒ \(x + z = 4\)
Add first and third: \(2x + 2y = 6\) ⇒ \(x + y = 3\)
From second: \(x – y + z = 2\) ⇒ \((x+z) – y = 2\) ⇒ \(4 – y = 2\) ⇒ \(y = 2\)
Then \(x = 1\), \(z = 3\)
Add first and third: \(2x + 2y = 6\) ⇒ \(x + y = 3\)
From second: \(x – y + z = 2\) ⇒ \((x+z) – y = 2\) ⇒ \(4 – y = 2\) ⇒ \(y = 2\)
Then \(x = 1\), \(z = 3\)
✅ \(x = 1, y = 2, z = 3\)
Q13 Find the condition for consistency: \(x + 2y = 3\), \(2x + 4y = k\).
Second equation is 2 × first equation ⇒ \(k = 2 \times 3 = 6\) for consistency.
✅ \(k = 6\)
Q14 Solve using inverse matrix: \(x + y = 2\), \(2x – y = 1\).
\(A = \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}\), \(A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 & -1 \\ -2 & 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}\)
\(X = A^{-1}B = \frac{1}{3} \begin{bmatrix} 1(2) + 1(1) \\ 2(2) + (-1)(1) \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
\(X = A^{-1}B = \frac{1}{3} \begin{bmatrix} 1(2) + 1(1) \\ 2(2) + (-1)(1) \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
✅ \(x = 1, y = 1\)
Q15 For what value of λ does the system have infinite solutions? \(x + y + z = 6\), \(2x + 3y + λz = 16\), \(x + 2y + 3z = 10\).
Using consistency conditions, after elimination we get \(λ = 4\) for infinite solutions.
✅ \(λ = 4\)
✏️ Practice Questions
P1 Solve: \(3x + y = 7\), \(x – 2y = 0\)
✅ Answer: \(x = 2, y = 1\)
P2 Check consistency: \(x + 3y = 5\), \(2x + 6y = 12\)
✅ Answer: Inconsistent (no solution)
P3 Using Cramer’s Rule: \(4x + y = 7\), \(2x – 3y = 0\)
✅ Answer: \(x = 1.5, y = 1\)
P4 Solve: \(x + 2y – z = 2\), \(2x + y + z = 5\), \(-x + y + z = 2\)
✅ Answer: \(x = 1, y = 1, z = 2\)
P5 Find k for unique solution: \(kx + 2y = 3\), \(x – y = 1\)
✅ Answer: \(k \neq -2\)
📋 Multiple Choice Questions (HSEB Pattern)
MCQ 1 A system of linear equations has a unique solution if:
A) \(D = 0\) B) \(D \neq 0\) C) \(D_x = 0\) D) \(D_y = 0\)
✅ Answer: B) \(D \neq 0\)
MCQ 2 The system \(x + 2y = 3\), \(2x + 4y = 6\) has:
A) Unique solution B) No solution C) Infinite solutions D) None
✅ Answer: C) Infinite solutions
MCQ 3 The method of solving linear equations using determinants is called:
A) Gaussian elimination B) Cramer’s Rule C) Matrix method D) Substitution
✅ Answer: B) Cramer’s Rule
MCQ 4 For a homogeneous system \(AX = 0\), the trivial solution exists when:
A) \(D = 0\) B) \(D \neq 0\) C) Always D) Never
✅ Answer: C) Always
MCQ 5 The system \(x + y = 5\), \(2x + 2y = 9\) is:
A) Consistent B) Inconsistent C) Dependent D) None
✅ Answer: B) Inconsistent
MCQ 6 In Gaussian elimination, the augmented matrix is reduced to:
A) Diagonal form B) Row echelon form C) Identity matrix D) Zero matrix
✅ Answer: B) Row echelon form
MCQ 7 If \(D = 0\) and \(D_x = 0\), the system has:
A) Unique solution B) No solution C) Infinite solutions or no solution D) None
✅ Answer: C) Infinite solutions or no solution
MCQ 8 The matrix form of a linear system is:
A) \(AX = B\) B) \(XA = B\) C) \(X = AB\) D) \(A = XB\)
✅ Answer: A) \(AX = B\)
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This PDF provides the solutions of every question from the exercises of class 12 system of linear equations chapter. If you want the notes of other exercises then you can choose from the button given above.
This PDF provides the solutions of every question from the exercises of class 12 system of linear equations chapter. If you want the notes of other exercises then you can choose from the button given above.
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📌 Key Formulas Summary:
• Cramer’s Rule (2 variables): \(x = \frac{D_x}{D}, y = \frac{D_y}{D}\)
• Matrix Solution: \(X = A^{-1}B\)
• Consistency Condition: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) → infinite; \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → no solution
• Gaussian Elimination: Reduce to row echelon form then back substitute
• Cramer’s Rule (2 variables): \(x = \frac{D_x}{D}, y = \frac{D_y}{D}\)
• Matrix Solution: \(X = A^{-1}B\)
• Consistency Condition: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) → infinite; \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) → no solution
• Gaussian Elimination: Reduce to row echelon form then back substitute