Cramer's Rule: Complete Case Analysis for JEE

Introduction

Cramer's Rule is a determinant-based method for solving systems of linear equations. Beyond providing solutions, it offers a systematic way to analyze the existence and uniqueness of solutions—a crucial skill for JEE.

Part I: Basic Setup

Consider a system of three linear equations: $\begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned}$

Key Determinants

1. Coefficient determinant ($\Delta$): $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

2. Variable determinants: $\Delta_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}, \quad \Delta_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}, \quad \Delta_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$

3. Solution (when $\Delta \neq 0$): $x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}$

Part II: Classification of Solutions

2.1 Non-Homogeneous Systems (at least one $d_i \neq 0$)

Explanation:

• Unique solution: Coefficient matrix is invertible ($\text{rank}(A) = 3$).
• Infinite solutions: System is consistent but dependent ($\text{rank}(A) = \text{rank}([A|B]) < 3$).
• No solution: System is inconsistent ($\text{rank}(A) < \text{rank}([A|B])$).

2.2 Homogeneous Systems ($d_1 = d_2 = d_3 = 0$)

• Always consistent (trivial solution $(0,0,0)$ always exists).
• Non-trivial solutions exist if and only if $\Delta = 0$.
• When $\Delta = 0$, there are infinitely many non-trivial solutions.

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Part III: Two-Variable Systems

For: $\begin{aligned} a_1x + b_1y &= c_1 \\ a_2x + b_2y &= c_2 \end{aligned}$ with $\Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}, \quad \Delta_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}, \quad \Delta_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$

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Part IV: JEE Previous Year Questions

PYQ 1 (JEE Main 2020)

The system: $\begin{aligned} x + y + z &= 6 \\ x + 2y + 3z &= 10 \\ x + 2y + \lambda z &= \mu \end{aligned}$ has no solution for: (A) $\lambda = 3, \mu \neq 10$
(B) $\lambda \neq 3$
(C) $\lambda = 3, \mu = 10$
(D) $\lambda \neq 3, \mu = 10$

Solution: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} = \lambda - 3$ For no solution: $\Delta = 0$ and at least one $\Delta_i \neq 0$.
$\Delta = 0 \implies \lambda = 3$.
When $\lambda = 3$, $\Delta_z = \mu - 10$. For $\Delta_z \neq 0$, $\mu \neq 10$.
Thus, no solution for $\lambda = 3, \mu \neq 10$.

Answer: (A)

PYQ 2 (JEE Main 2021)

Let S be the set of all $\lambda \in \mathbb{R}$ for which the system: $\begin{aligned} 2x - y + 2z &= 2 \\ x - 2y + \lambda z &= -4 \\ x + \lambda y + z &= 4 \end{aligned}$ has no solution. Then S is: (A) empty (B) a singleton (C) contains exactly two elements (D) contains more than two elements

Solution: $\Delta = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{vmatrix} = -2\lambda^2 + \lambda + 1 = -(2\lambda+1)(\lambda-1)$ For no solution: $\Delta = 0$ and at least one $\Delta_i \neq 0$.
$\Delta = 0 \implies \lambda = 1$ or $\lambda = -\frac{1}{2}$.
For both values, $\Delta_x \neq 0$, so both give no solution.
Thus, S has two elements.

Answer: (C)

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Part V: Quick Reference Summary

Decision Flowchart

1. Compute $\Delta$.
2. If $\Delta \neq 0$: unique solution exists.
3. If $\Delta = 0$: compute $\Delta_x, \Delta_y, \Delta_z$.
   • If all $\Delta_i = 0$: infinite solutions.
   • If at least one $\Delta_i \neq 0$: no solution.

Homogeneous Systems

• Always consistent.
• $\Delta \neq 0$: only trivial solution.
• $\Delta = 0$: infinite non-trivial solutions.

Common Mistakes

• Forgetting to check all $\Delta_i$ when $\Delta = 0$.
• Assuming homogeneous systems can have no solution.
• Confusing "infinite solutions" with "no solution" when $\Delta = 0$.

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Practice Problems

1. For what value of $k$ does the system
   $x + 2y + 3z = 1$, $2x + 4y + 6z = k$, $3x + 6y + 9z = 3$ have infinite solutions?
2. Determine the condition for the homogeneous system
   $ax + by + cz = 0$, $bx + cy + az = 0$, $cx + ay + bz = 0$ to have non-trivial solutions.
3. Find $\lambda$ such that the system
   $x + y + z = 6$, $x + 2y + 3z = 10$, $x + 2y + \lambda z = 12$ has a unique solution.

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Key Takeaway: Cramer's Rule provides both a solution method and a complete classification tool for linear systems. Mastering its cases is essential for efficiently tackling JEE problems on systems of equations.