Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers Matrix Inversion Method and Cramer's Rule for solving systems of linear equations, with applications in electrical circuits, forces, beams, 2D game worlds, ages, cyclists, and fractions. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 3: Matrices and Determinants
Class 10 Math Unit 3 Exercise 3.6 – Matrix Inversion & Cramer's Rule: Complete Guide
Exercise 3.6 of Unit 3 introduces two powerful methods for solving systems of linear equations: the Matrix Inversion Method (\( X = A^{-1}B \)) and Cramer's Rule (\( x = |A_x|/|A| \), \( y = |A_y|/|A| \)). These methods are essential for solving real-world problems in electrical circuits, structural engineering, game development, and more.
What You Will Learn
By working through Exercise 3.6, students master: solving systems of linear equations using matrix inversion (finding \( A^{-1} \) and applying \( X = A^{-1}B \)), applying Cramer's rule using determinants (\( |A| \), \( |A_x| \), \( |A_y| \)), and solving word problems involving electrical circuits, forces, support beams, 2D game worlds, ages, cyclists, and fractions.
Topics Covered in This Exercise
- Matrix Inversion Method – solving \( AX = B \) using \( X = A^{-1}B \)
- Cramer's Rule – using determinants to solve \( 2 \times 2 \) systems
- Electrical Circuit – finding current in two branches
- Forces in Equilibrium – solving force vector equations
- Support Beams – determining load distribution
- 2D Game World – finding point of intersection
- Ages Problem – solving age-related word problems
- Cyclists Problem – finding speeds using relative motion
- Fraction Problem – finding original fraction from conditions
Why Exercise 3.6 Is Important
Exercise 3.6 is the culmination of Unit 3. Board exam papers frequently feature questions on matrix inversion and Cramer's rule in both pure mathematical and word problem formats. This exercise demonstrates how matrices are used in real-world applications, making it one of the most important exercises in the unit.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Matrix inversion and Cramer's rule are frequently tested topics, and word problems appear regularly in board papers. Practising every part of Q1, Q2, and the application problems (Q3-Q9) by hand is highly recommended.
Exam Tips for Matrix Inversion & Cramer's Rule
- Always check that \( |A| \neq 0 \) before using either method.
- For matrix inversion, show all steps: \( A^{-1} = \frac{1}{|A|}\text{Adj}(A) \).
- For Cramer's rule, clearly define \( A_x \) and \( A_y \) matrices.
- In word problems, identify the variables and write the system first.
Common Mistakes Students Make
- Forgetting to calculate \( |A| \) and checking if it's non-zero.
- Writing the adjugate matrix incorrectly (sign errors).
- Confusing \( A_x \) and \( A_y \) in Cramer's rule.
- Not simplifying final answers properly.
1
Solve by matrix inversion method, if possible
(i) \(2x + 5y = 19\) ; \(4x - 3y = -1\)
(ii) \(3x + 2y = 7\) ; \(5x - y = 16\)
(iii) \(x - 2y = 9\) ; \(2x + 7y = -4\)
(iv) \(3x + 2y = 2\) ; \(x - 2y = -2\)
(i) \(2x+5y=19\) ; \(4x-3y=-1\)
In matrix form
\[ \begin{bmatrix} 2 & 5 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 19 \\ -1 \end{bmatrix} \]Let \(AX = B\), \(X = A^{-1}B\) ... (i)
Now
\[ A = \begin{bmatrix} 2 & 5 \\ 4 & -3 \end{bmatrix} \] \[ |A| = (2)(-3) - (5)(4) = -6 - 20 = -26 \]\(|A| \neq 0\), A is non singular matrix, so \(A^{-1}\) exists
\[ \operatorname{Adj} A = \begin{bmatrix} -3 & -5 \\ -4 & 2 \end{bmatrix} \] \[ A^{-1} = \frac{1}{-26} \begin{bmatrix} -3 & -5 \\ -4 & 2 \end{bmatrix} \]From equation (i)
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-26} \begin{bmatrix} -3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 19 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]Thus \(x=2\) and \(y=3\)
\(S.S = \{(2,3)\}\)
(ii) \(3x+2y=7\) ; \(5x-y=16\)
In matrix form
\[ \begin{bmatrix} 3 & 2 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 16 \end{bmatrix} \]Let \(AX=B\), \(X=A^{-1}B\) ... (i)
Now
\[ A = \begin{bmatrix} 3 & 2 \\ 5 & -1 \end{bmatrix} \] \[ |A| = (3)(-1) - (2)(5) = -3 - 10 = -13 \]\(|A| \neq 0\), so \(A^{-1}\) exists
\[ \operatorname{Adj} A = \begin{bmatrix} -1 & -2 \\ -5 & 3 \end{bmatrix} \] \[ A^{-1} = \frac{1}{-13} \begin{bmatrix} -1 & -2 \\ -5 & 3 \end{bmatrix} \]From equation (i)
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-13} \begin{bmatrix} -1 & -2 \\ -5 & 3 \end{bmatrix} \begin{bmatrix} 7 \\ 16 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \]Thus \(x=3\) and \(y=-1\)
\(S.S = \{(3,-1)\}\)
(iii) \(x-2y=9\) ; \(2x+7y=-4\)
In matrix form
\[ \begin{bmatrix} 1 & -2 \\ 2 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9 \\ -4 \end{bmatrix} \]Let \(AX=B\), \(X=A^{-1}B\) ... (i)
Now
\[ A = \begin{bmatrix} 1 & -2 \\ 2 & 7 \end{bmatrix}, \quad |A| = (1)(7) - (-2)(2) = 7 + 4 = 11 \]\(|A| \neq 0\), so \(A^{-1}\) exists
\[ \operatorname{Adj} A = \begin{bmatrix} 7 & 2 \\ -2 & 1 \end{bmatrix} \] \[ A^{-1} = \frac{1}{11} \begin{bmatrix} 7 & 2 \\ -2 & 1 \end{bmatrix} \]From (i)
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 7 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 9 \\ -4 \end{bmatrix} = \begin{bmatrix} 5 \\ -2 \end{bmatrix} \]Thus \(x=5\) and \(y=-2\)
\(S.S = \{(5,-2)\}\)
(iv) \(3x+2y=2\) ; \(x-2y=-2\)
In matrix form
\[ \begin{bmatrix} 3 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix} \]Let \(AX=B\), \(X=A^{-1}B\) ... (i)
Now
\[ A = \begin{bmatrix} 3 & 2 \\ 1 & -2 \end{bmatrix}, \quad |A| = (3)(-2) - (2)(1) = -6 - 2 = -8 \]\(|A| \neq 0\), so \(A^{-1}\) exists
\[ \operatorname{Adj} A = \begin{bmatrix} -2 & -2 \\ -1 & 3 \end{bmatrix} \] \[ A^{-1} = \frac{1}{-8} \begin{bmatrix} -2 & -2 \\ -1 & 3 \end{bmatrix} \]From (i)
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-8} \begin{bmatrix} -2 & -2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]Thus \(x=0\) and \(y=1\)
\(S.S = \{(0,1)\}\)
2
Use Cramer's rule to solve the following pair of linear equations, if possible
(i) \(x+4y=4\) ; \(2x-y=5\)
(ii) \(x+2y=7\) ; \(3x-2y=1\)
(iii) \(2x-5y=-6\) ; \(4x-5y=-2\)
(iv) \(3x+2y=-1\) ; \(5x+6y=5\)
(i) \(x+4y=4\) ; \(2x-y=5\)
In matrix form
\[ \begin{bmatrix} 1 & 4 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & 4 \\ 2 & -1 \end{bmatrix}, \quad A_x = \begin{bmatrix} 4 & 4 \\ 5 & -1 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix} \] \[ |A| = -9,\quad |A_x| = -24,\quad |A_y| = -3 \] \[ x = \frac{|A_x|}{|A|} = \frac{-24}{-9} = \frac{8}{3},\quad y = \frac{|A_y|}{|A|} = \frac{-3}{-9} = \frac{1}{3} \]Hence \(S.S = \left\{\left(\frac{8}{3}, \frac{1}{3}\right)\right\}\)
(ii) \(x+2y=7\) ; \(3x-2y=1\)
In matrix form
\[ \begin{bmatrix} 1 & 2 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & 2 \\ 3 & -2 \end{bmatrix}, \quad A_x = \begin{bmatrix} 7 & 2 \\ 1 & -2 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & 7 \\ 3 & 1 \end{bmatrix} \] \[ |A| = -8,\quad |A_x| = -16,\quad |A_y| = -20 \] \[ x = \frac{-16}{-8} = 2,\quad y = \frac{-20}{-8} = \frac{5}{2} \]Hence \(S.S = \{(2, \frac{5}{2})\}\)
(iii) \(2x-5y=-6\) ; \(4x-5y=-2\)
In matrix form
\[ \begin{bmatrix} 2 & -5 \\ 4 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -6 \\ -2 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 2 & -5 \\ 4 & -5 \end{bmatrix}, \quad A_x = \begin{bmatrix} -6 & -5 \\ -2 & -5 \end{bmatrix}, \quad A_y = \begin{bmatrix} 2 & -6 \\ 4 & -2 \end{bmatrix} \] \[ |A| = 10,\quad |A_x| = 20,\quad |A_y| = 20 \] \[ x = \frac{20}{10} = 2,\quad y = \frac{20}{10} = 2 \]Hence \(S.S = \{(2,2)\}\)
(iv) \(3x+2y=-1\) ; \(5x+6y=5\)
In matrix form
\[ \begin{bmatrix} 3 & 2 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 \\ 5 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 3 & 2 \\ 5 & 6 \end{bmatrix}, \quad A_x = \begin{bmatrix} -1 & 2 \\ 5 & 6 \end{bmatrix}, \quad A_y = \begin{bmatrix} 3 & -1 \\ 5 & 5 \end{bmatrix} \] \[ |A| = 8,\quad |A_x| = -16,\quad |A_y| = 20 \] \[ x = \frac{-16}{8} = -2,\quad y = \frac{20}{8} = \frac{5}{2} \]Hence \(S.S = \left\{\left(-2, \frac{5}{2}\right)\right\}\)
3
Electrical circuit – current in branches
An electrical engineer wants to determine the current in two branches \(A\) and \(B\) of a simple electrical circuit. The system of the equations is:
\[ x+y=7 \] \[ 2x-y=2 \]
where \(x\) is the current in branch \(A\) and \(y\) is the current in branch \(B\). Find \(x\) and \(y\) by using matrices.
\(x+y=7\)
\(2x-y=2\)
In matrix form
\[ \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 2 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}, \quad A_x = \begin{bmatrix} 7 & 1 \\ 2 & -1 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & 7 \\ 2 & 2 \end{bmatrix} \] \[ |A| = -3,\quad |A_x| = -9,\quad |A_y| = -12 \] \[ x = \frac{-9}{-3} = 3,\quad y = \frac{-12}{-3} = 4 \]Hence Current in branch \(A = x = 3\), Current in branch \(B = y = 4\)
4
Forces in equilibrium
Three forces act on a particle and must be in equilibrium i.e. \(F_1 + F_2 + F_3 = 0\), where \(F_1 = \begin{bmatrix} 8 \\ x \end{bmatrix}\), \(F_2 = \begin{bmatrix} -2 \\ -7 \end{bmatrix}\), \(F_3 = \begin{bmatrix} y \\ -1 \end{bmatrix}\). Find the value of \(x\) and \(y\).
Since \(F_1 + F_2 + F_3 = 0\)
\[ \begin{bmatrix} 8 \\ x \end{bmatrix} + \begin{bmatrix} -2 \\ -7 \end{bmatrix} + \begin{bmatrix} y \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] \[ \begin{bmatrix} 6+y \\ x-8 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]By comparing: \(6+y=0 \Rightarrow y=-6\), \(x-8=0 \Rightarrow x=8\)
Hence \(x=8\), \(y=-6\)
5
Support beams load
Two support beams, \(A\) and \(B\) are holding up a combined load of \(100\,kN\). Twice the load on beam A and three times the load on beam B equals \(240\,kN\). Find the load of beam A and beam B by using matrices.
Let load on beam \(A = x\,kN\), load on beam \(B = y\,kN\)
According to first condition: \(x+y=100\)
According to second condition: \(2x+3y=240\)
In matrix form
\[ \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 100 \\ 240 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}, \quad A_x = \begin{bmatrix} 100 & 1 \\ 240 & 3 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & 100 \\ 2 & 240 \end{bmatrix} \] \[ |A| = 1,\quad |A_x| = 60,\quad |A_y| = 40 \] \[ x = \frac{60}{1} = 60,\quad y = \frac{40}{1} = 40 \]Hence Load on beam \(A = 60\,kN\), Load on beam \(B = 40\,kN\)
6
2D game world – point of intersection
In a 2D game world, two characters are moving along straight paths. One character moves along a line where the total of twice their horizontal position and vertical position is 5, while the other moves along a line where their horizontal position is one more than their vertical position. Find their point of intersection by using matrices.
Let horizontal position \(= x\), vertical position \(= y\)
According to first condition: \(2x+y=5\)
According to second condition: \(x = y+1 \Rightarrow x-y = 1\)
In matrix form
\[ \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}, \quad A_x = \begin{bmatrix} 5 & 1 \\ 1 & -1 \end{bmatrix}, \quad A_y = \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix} \] \[ |A| = -3,\quad |A_x| = -6,\quad |A_y| = -3 \] \[ x = \frac{-6}{-3} = 2,\quad y = \frac{-3}{-3} = 1 \]Hence point of intersection \((x,y) = (2,1)\)
7
Present ages (father & son)
Two years ago a man was 5 times as old as his son was. After 6 years he will be 3 times as old as his son. Find their present ages by using matrices.
Let present age of man \(= x\) years, present age of son \(= y\) years
According to first condition (two years ago): \(x-2 = 5(y-2) \Rightarrow x - 5y = -8\)
According to second condition (after 6 years): \(x+6 = 3(y+6) \Rightarrow x - 3y = 12\)
In matrix form
\[ \begin{bmatrix} 1 & -5 \\ 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -8 \\ 12 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & -5 \\ 1 & -3 \end{bmatrix}, \quad A_x = \begin{bmatrix} -8 & -5 \\ 12 & -3 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & -8 \\ 1 & 12 \end{bmatrix} \] \[ |A| = 2,\quad |A_x| = 84,\quad |A_y| = 20 \] \[ x = \frac{84}{2} = 42,\quad y = \frac{20}{2} = 10 \]Hence present age of man \(= 42\) years, son \(= 10\) years.
8
Two cyclists – speeds
Two cyclists are 44 km apart and start out at the same time. If they go towards one another they meet in 2 hours, but if they go in the same direction the faster overtakes the slower in \(7\frac{1}{2}\) hours. Find their speeds by using matrices.
Let speed of faster cyclist \(= x\,kmh^{-1}\), speed of slower cyclist \(= y\,kmh^{-1}\)
According to first condition (moving towards one another):
\(44 = (x+y) \times 2 \Rightarrow x+y = 22\)
According to second condition (same direction, faster overtakes):
\(44 = (x-y) \times \frac{15}{2} \Rightarrow 15x - 15y = 88\)
In matrix form
\[ \begin{bmatrix} 1 & 1 \\ 15 & -15 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 22 \\ 88 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & 1 \\ 15 & -15 \end{bmatrix}, \quad A_x = \begin{bmatrix} 22 & 1 \\ 88 & -15 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & 22 \\ 15 & 88 \end{bmatrix} \] \[ |A| = -30,\quad |A_x| = -418,\quad |A_y| = -242 \] \[ x = \frac{-418}{-30} = \frac{209}{15},\quad y = \frac{-242}{-30} = \frac{121}{15} \]Hence speeds: faster \(= \frac{209}{15}\,kmh^{-1}\), slower \(= \frac{121}{15}\,kmh^{-1}\)
9
Fraction problem
The numerator of a fraction is 7 less than the denominator. If the numerator is increased by 3, the new fraction can be cancelled down to \(\frac{3}{4}\). Find the original fraction by using matrices.
Let numerator \(= x\), denominator \(= y\). Original fraction \(= \frac{x}{y}\)
According to first condition: \(x = y-7 \Rightarrow x - y = -7\)
According to second condition: \(\frac{x+3}{y} = \frac{3}{4} \Rightarrow 4x - 3y = -12\)
In matrix form
\[ \begin{bmatrix} 1 & -1 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ -12 \end{bmatrix} \]Let
\[ A = \begin{bmatrix} 1 & -1 \\ 4 & -3 \end{bmatrix}, \quad A_x = \begin{bmatrix} -7 & -1 \\ -12 & -3 \end{bmatrix}, \quad A_y = \begin{bmatrix} 1 & -7 \\ 4 & -12 \end{bmatrix} \] \[ |A| = 1,\quad |A_x| = 9,\quad |A_y| = 16 \] \[ x = \frac{9}{1} = 9,\quad y = \frac{16}{1} = 16 \]Hence original fraction \(= \frac{9}{16}\)
📖 Multiple Choice Questions (Unit 3 Review)
1. For the matrix inversion method, \( X = A^{-1}B \) is valid when:
✅ Correct Answer: (B) \( |A| \neq 0 \)
Matrix inversion method requires the coefficient matrix \( A \) to be non-singular, i.e., \( |A| \neq 0 \).
Matrix inversion method requires the coefficient matrix \( A \) to be non-singular, i.e., \( |A| \neq 0 \).
2. In Cramer's rule, \( x = \frac{|A_x|}{|A|} \). What is \( A_x \)?
✅ Correct Answer: (A) Matrix \( A \) with first column replaced by \( B \)
\( A_x \) is formed by replacing the first column (coefficients of \( x \)) of matrix \( A \) with the constant matrix \( B \).
\( A_x \) is formed by replacing the first column (coefficients of \( x \)) of matrix \( A \) with the constant matrix \( B \).
3. For the system \( x + y = 7 \), \( 2x - y = 2 \), what is \( |A| \)?
✅ Correct Answer: (A) \(-3\)
\( A = \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} \), \( |A| = (1)(-1) - (1)(2) = -1 - 2 = -3 \).
\( A = \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} \), \( |A| = (1)(-1) - (1)(2) = -1 - 2 = -3 \).
4. The adjugate of \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is:
✅ Correct Answer: (A) \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
The adjugate of a \( 2 \times 2 \) matrix swaps the diagonal elements and changes the sign of the off-diagonal elements.
The adjugate of a \( 2 \times 2 \) matrix swaps the diagonal elements and changes the sign of the off-diagonal elements.
5. In Cramer's rule, \( y = \frac{|A_y|}{|A|} \). What is \( A_y \) for the system \( 2x+y=5 \), \( x-y=1 \)?
✅ Correct Answer: (A) \( \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix} \)
\( A_y \) is formed by replacing the second column (coefficients of \( y \)) of matrix \( A \) with the constant matrix \( B \).
\( A_y \) is formed by replacing the second column (coefficients of \( y \)) of matrix \( A \) with the constant matrix \( B \).
📈 Key Concepts – Matrix Inversion & Cramer's Rule
- Matrix Inversion Method: \( X = A^{-1}B \), where \( A^{-1} = \frac{1}{|A|}\text{Adj}(A) \), requires \( |A| \neq 0 \)
- Cramer's Rule: \( x = \frac{|A_x|}{|A|} \), \( y = \frac{|A_y|}{|A|} \) for \( 2 \times 2 \) systems, requires \( |A| \neq 0 \)
- Adjugate of \( 2 \times 2 \) matrix: \( \text{Adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
- Applications: Electrical circuits, forces, beams, game worlds, ages, cyclists, fractions
❓ Frequently Asked Questions
What is taught in Exercise 3.6 of Class 10 Math Unit 3?
Exercise 3.6 covers two methods for solving systems of linear equations: Matrix Inversion Method (\( X = A^{-1}B \)) and Cramer's Rule (\( x = |A_x|/|A| \), \( y = |A_y|/|A| \)). It includes applications in electrical circuits, forces, support beams, 2D game worlds, ages, cyclists, and fractions.
What is the Matrix Inversion Method?
The Matrix Inversion Method solves \( AX = B \) by finding \( X = A^{-1}B \), where \( A^{-1} = \frac{1}{|A|} \text{Adj}(A) \). This method requires the determinant \( |A| \neq 0 \).
What is Cramer's Rule?
Cramer's Rule solves systems of linear equations using determinants. For a \( 2 \times 2 \) system, \( x = \frac{|A_x|}{|A|} \) and \( y = \frac{|A_y|}{|A|} \), where \( |A| \neq 0 \).
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 3.6 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 3.6?
Yes, a complete solved PDF for Exercise 3.6 is embedded on this page and available to download for free.
Can I download the Unit 3 Exercise 3.6 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 3.6 notes to your device.
Is Exercise 3.6 important for Class 10 board exams?
Yes, matrix inversion and Cramer's rule are frequently tested in board exams. Exercise 3.6 builds essential skills in solving systems of linear equations and their real-world applications.
Who prepared these Class 10 Math Unit 3 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.