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JEE Main & Advanced Reading Time: 20 min 2 Methods + Applications

Solving Systems of Equations using Matrices: Cramer's Rule & Matrix Method

Master matrix methods for solving linear systems with step-by-step examples and JEE-level practice problems.

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Powerful Methods
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JEE Relevance
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Solved Examples
30min
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Why Matrix Methods are Essential for JEE

Matrix methods provide systematic approaches to solve systems of linear equations, appearing frequently in JEE Main and Advanced. These techniques are crucial for:

  • Efficient solving of 2×2 and 3×3 systems in less time
  • Consistency analysis of linear systems
  • Geometric interpretations of solutions
  • Application problems in physics and coordinate geometry
Matrix Method Medium

Inverse Matrix Method

Solve $AX = B$ using $X = A^{-1}B$ where $A$ is the coefficient matrix.

When to Use:

When coefficient matrix $A$ is square and invertible ($|A| \neq 0$). Most efficient for 2×2 and 3×3 systems.

Example: Solve the system

$$2x + 3y = 8$$ $$4x - y = 2$$

Step 1: Write in matrix form $AX = B$:

$\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 2 \end{bmatrix}$

Step 2: Find $A^{-1}$:

$|A| = (2)(-1) - (3)(4) = -2 - 12 = -14$

$A^{-1} = \frac{1}{-14}\begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix}$

Step 3: Multiply $X = A^{-1}B$:

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} 8 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

Step 4: Solution: $x = 1$, $y = 2$

Cramer's Rule Hard

Cramer's Rule for n×n Systems

$x_i = \frac{|A_i|}{|A|}$ where $A_i$ is matrix $A$ with $i^{th}$ column replaced by $B$.

When to Use:

When you need specific variables or when $|A| \neq 0$. Particularly useful for 3×3 systems.

Example: Solve using Cramer's Rule

$$x + 2y + z = 7$$ $$2x - y + 3z = 9$$ $$3x + y - z = 1$$

Step 1: Calculate $|A|$:

$|A| = \begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & 3 \\ 3 & 1 & -1 \end{vmatrix} = 1(1-3) - 2(-2-9) + 1(2+3) = -2 + 22 + 5 = 25$

Step 2: Calculate $|A_1|$ (replace first column):

$|A_1| = \begin{vmatrix} 7 & 2 & 1 \\ 9 & -1 & 3 \\ 1 & 1 & -1 \end{vmatrix} = 7(1-3) - 2(-9-3) + 1(9+1) = -14 + 24 + 10 = 20$

Step 3: Calculate $|A_2|$ (replace second column):

$|A_2| = \begin{vmatrix} 1 & 7 & 1 \\ 2 & 9 & 3 \\ 3 & 1 & -1 \end{vmatrix} = 1(-9-3) - 7(-2-9) + 1(2-27) = -12 + 77 - 25 = 40$

Step 4: Calculate $|A_3|$ (replace third column):

$|A_3| = \begin{vmatrix} 1 & 2 & 7 \\ 2 & -1 & 9 \\ 3 & 1 & 1 \end{vmatrix} = 1(-1-9) - 2(2-27) + 7(2+3) = -10 + 50 + 35 = 75$

Step 5: Apply Cramer's Rule:

$x = \frac{|A_1|}{|A|} = \frac{20}{25} = \frac{4}{5}$

$y = \frac{|A_2|}{|A|} = \frac{40}{25} = \frac{8}{5}$

$z = \frac{|A_3|}{|A|} = \frac{75}{25} = 3$

🚀 Problem-Solving Strategies

Method Selection Guide:

  • 2×2 systems: Inverse method or Cramer's Rule
  • 3×3 systems: Cramer's Rule preferred
  • When |A| = 0: Check consistency using rank
  • For specific variable: Cramer's Rule

Common Pitfalls:

  • Forgetting to check if |A| ≠ 0
  • Sign errors in determinant calculation
  • Matrix multiplication order mistakes
  • Not verifying solutions in original equations
JEE Applications Advanced

Real JEE Problems & Applications

JEE Main 2023 Problem:

For what value of $k$ does the system have infinitely many solutions?

$$x + 2y + z = 3$$ $$2x + 3y + 2z = 5$$ $$4x + 7y + 4z = k$$

Step 1: Write augmented matrix and reduce:

$\begin{bmatrix} 1 & 2 & 1 & | & 3 \\ 2 & 3 & 2 & | & 5 \\ 4 & 7 & 4 & | & k \end{bmatrix}$

Step 2: For infinite solutions, rank(A) = rank([A|B]) < number of variables

Step 3: After row reduction, condition becomes $k = 9$

Coordinate Geometry Application:

Find the equation of plane passing through (1,2,3), (2,3,4), and (3,4,5)

Step 1: Let plane equation be $ax + by + cz + d = 0$

Step 2: Substitute points to get system:

$a + 2b + 3c + d = 0$

$2a + 3b + 4c + d = 0$

$3a + 4b + 5c + d = 0$

Step 3: Solve using matrix methods

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. Solve using matrix method: $3x - 2y = 7$, $x + 4y = 9$

2. Use Cramer's Rule: $2x + y - z = 3$, $x - y + z = 1$, $3x + 2y + z = 4$

3. Find $k$ for consistent system: $x + y + z = 1$, $2x + 3y + kz = 4$, $x + ky + 3z = 1$

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