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IIT JEE Main Maths -Unit 3- Consistency and solution of linear equations (2 or 3 variables)- Study Notes-New Syllabus

IIT JEE Main Maths -Unit 3- Consistency and solution of linear equations (2 or 3 variables) – Study Notes – New syllabus

IIT JEE Main Maths -Unit 3- Consistency and solution of linear equations (2 or 3 variables) – Study Notes -IIT JEE Main Maths – per latest Syllabus.

Key Concepts:

  • Applications of Determinants and Matrices
  • Properties and Applications of Inverse Matrix — Orthogonal and Singular Transformations

IIT JEE Main Maths -Study Notes – All Topics

Applications of Determinants and Matrices — Consistency and Solving System of Linear Equations (Cramer’s Rule & Matrix Method)

Introduction

Determinants and matrices are powerful tools for solving systems of linear equations of the 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 $

This system can be represented in matrix form as:

$ A X = B $

where

  • \( A = \begin{bmatrix}a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{bmatrix} \) — coefficient matrix
  • \( X = \begin{bmatrix}x\\y\\z\end{bmatrix} \) — variable matrix
  • \( B = \begin{bmatrix}d_1\\d_2\\d_3\end{bmatrix} \) — constant matrix

Consistency of a System

Let \( A X = B \) represent a system of equations.

  • If \( |A| \ne 0 \), the system is consistent and has a unique solution.
  • If \( |A| = 0 \), the system may be consistent (infinite solutions) or inconsistent (no solution) depending on \( \text{adj}(A)B = 0 \) or not.

Condition Summary:

CaseConditionNature of Solutions
1\( |A| \ne 0 \)Unique solution (consistent)
2\( |A| = 0 \) and \( \text{adj}(A)B = 0 \)Infinitely many solutions (consistent)
3\( |A| = 0 \) and \( \text{adj}(A)B \ne 0 \)No solution (inconsistent)

 Cramer’s Rule (for 2 or 3 Variables)

If \( |A| \ne 0 \), the system has a unique solution given by:

$ x = \dfrac{|A_x|}{|A|}, \quad y = \dfrac{|A_y|}{|A|}, \quad z = \dfrac{|A_z|}{|A|} $

where \( A_x, A_y, A_z \) are obtained by replacing the 1st, 2nd, and 3rd columns of \( A \) with \( B \), respectively.

Matrix Method

For \( A X = B \):

If \( |A| \ne 0 \), then multiplying both sides by \( A^{-1} \):

\( A^{-1} A X = A^{-1} B \Rightarrow X = A^{-1} B \)

This gives a compact and powerful method to find \( x, y, z \).

Example

Solve the equations using Cramer’s rule:

$ 2x + 3y = 8\\ x + 2y = 5 $

▶️ Answer / Explanation

\( |A| = \begin{vmatrix}2 & 3\\1 & 2\end{vmatrix} = 4 – 3 = 1 \)

\( |A_x| = \begin{vmatrix}8 & 3\\5 & 2\end{vmatrix} = 16 – 15 = 1 \)

\( |A_y| = \begin{vmatrix}2 & 8\\1 & 5\end{vmatrix} = 10 – 8 = 2 \)

\( x = \dfrac{1}{1} = 1, \quad y = \dfrac{2}{1} = 2 \)

Answer: \( x = 1, y = 2 \)

Example 

Solve the system using matrix method:

$ x + 2y + 3z = 10\\ 2x + 3y + z = 8\\ 3x + y + 2z = 9 $

▶️ Answer / Explanation

\( A = \begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\\3 & 1 & 2\end{bmatrix}, \quad B = \begin{bmatrix}10\\8\\9\end{bmatrix} \)

\( |A| = 1(3×2 – 1×1) – 2(2×2 – 1×3) + 3(2×1 – 3×3) = 1(5) – 2(1) + 3(-7) = 5 – 2 – 21 = -18 \)

Find \( A^{-1} \) using adjoint (skipped steps for brevity): \( A^{-1} = \dfrac{1}{-18}\begin{bmatrix}5 & -1 & -7\\-4 & -7 & 11\\2 & 13 & -11\end{bmatrix} \)

\( X = A^{-1}B = \dfrac{1}{-18} \begin{bmatrix}5 & -1 & -7\\-4 & -7 & 11\\2 & 13 & -11\end{bmatrix} \begin{bmatrix}10\\8\\9\end{bmatrix} = \begin{bmatrix}1\\2\\3\end{bmatrix} \)

Answer: \( x = 1, y = 2, z = 3 \)

Example 

Examine the consistency of the system:

$ x + y + z = 6\\ 2x + 3y + z = 10\\ 3x + 4y + 2z = 14 $

▶️ Answer / Explanation

\( A = \begin{bmatrix}1 & 1 & 1\\2 & 3 & 1\\3 & 4 & 2\end{bmatrix}, \quad B = \begin{bmatrix}6\\10\\14\end{bmatrix} \)

\( |A| = 1(3×2 – 1×4) – 1(2×2 – 1×3) + 1(2×4 – 3×3) = (2) – (1) + (-1) = 0 \)

Since \( |A| = 0 \), check \( \text{adj}(A)B \): After computation, \( \text{adj}(A)B = 0 \).

Hence, the system is consistent and has infinitely many solutions.


Key Takeaways for JEE:

  • Use Cramer’s Rule for small systems (2×2 or 3×3).
  • Matrix method \( X = A^{-1}B \) is compact and direct for unique solutions.
  • Always check \( |A| \) first for consistency.
  • If \( |A| = 0 \) and \( \text{adj}(A)B \ne 0 \), → No solution (inconsistent).
  • If \( |A| = 0 \) and \( \text{adj}(A)B = 0 \), → Infinite solutions (consistent).

Properties and Applications of Inverse Matrix — Orthogonal and Singular Transformations

 Recap — Inverse of a Matrix

If \( A \) is a square matrix and \( |A| \ne 0 \), its inverse exists and is given by:

\( A^{-1} = \dfrac{1}{|A|} \text{adj}(A) \)

The inverse satisfies the relation:

\( A A^{-1} = A^{-1} A = I \)

 Singular and Non-Singular Matrices

TypeConditionInverse Exists?
Non-Singular Matrix\( |A| \ne 0 \)Yes
Singular Matrix\( |A| = 0 \)No

 Orthogonal Matrix

A square matrix \( A \) is called orthogonal if:

\( A^T A = A A^T = I \)

Hence, \( A^{-1} = A^T \)

Properties of Orthogonal Matrices:

  • All orthogonal matrices are non-singular.
  • \( |A| = \pm 1 \).
  • The product of two orthogonal matrices is orthogonal.
  • The transpose of an orthogonal matrix is also orthogonal.

 Applications of Inverse in Linear Transformations

Matrices can represent linear transformations like rotation, reflection, and scaling. If \( T \) is a transformation represented by matrix \( A \), then the inverse transformation \( T^{-1} \) is represented by \( A^{-1} \).

Example:

If a rotation in 2D space is represented by $ R = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}, $ then

$ R^{-1} = R^T = \begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} $ represents the inverse (rotation by \(-\theta\)).

Properties of Inverse Matrices (Extended)

PropertyStatement
1\( (A^{-1})^{-1} = A \)
2\( (AB)^{-1} = B^{-1}A^{-1} \)
3\( (A^T)^{-1} = (A^{-1})^T \)
4If \( A \) is orthogonal, \( A^{-1} = A^T \)
5If \( A \) is invertible, then \( (kA)^{-1} = \dfrac{1}{k}A^{-1} \)

Example

Show that \( A = \begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix} \) is orthogonal.

▶️ Answer / Explanation

Compute \( A^T A \):

$ A^T = \begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix} $

$ A^T A = \begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix} = \begin{bmatrix}\cos^2\theta + \sin^2\theta & 0\\0 & \cos^2\theta + \sin^2\theta\end{bmatrix} = I $

Hence proved: \( A \) is orthogonal and \( A^{-1} = A^T \).

Example 

Find \( A^{-1} \) if \( A = \begin{bmatrix}2 & 1\\7 & 4\end{bmatrix} \) and verify \( A A^{-1} = I \).

▶️ Answer / Explanation

\( |A| = 2(4) – 7(1) = 8 – 7 = 1 \)

\( \text{adj}(A) = \begin{bmatrix}4 & -1\\-7 & 2\end{bmatrix} \)

\( A^{-1} = \dfrac{1}{1} \begin{bmatrix}4 & -1\\-7 & 2\end{bmatrix} = \begin{bmatrix}4 & -1\\-7 & 2\end{bmatrix} \)

Check: $ A A^{-1} = \begin{bmatrix}2 & 1\\7 & 4\end{bmatrix}\begin{bmatrix}4 & -1\\-7 & 2\end{bmatrix} = \begin{bmatrix}8 – 7 & -2 + 2\\28 – 28 & -7 + 8\end{bmatrix} = I $

Hence verified.

Example 

Given that a transformation matrix \( T = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix} \) rotates a vector 90° anticlockwise, find \( T^{-1} \) and explain its effect.

▶️ Answer / Explanation

\( T^T = \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} \)

\( T^T T = I \Rightarrow T^{-1} = T^T \)

\( T^{-1} = \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix} \) represents a rotation by −90° (clockwise).

Conclusion: The inverse of a rotation matrix rotates by the same angle in the opposite direction.

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