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.
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
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 $AX = B$:
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 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
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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