The Hook: GPS Uses Linear Equations Every Second
Every time you open Google Maps, your phone solves a system of linear equations in milliseconds!
GPS satellites send signals, and your phone needs to find its position $(x, y, z)$ by solving:
$$\begin{cases} (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = d_1^2 \\ (x - x_2)^2 + (y - y_2)^2 + (z - z_2)^2 = d_2^2 \\ (x - x_3)^2 + (y - y_3)^2 + (z - z_3)^2 = d_3^2 \end{cases}$$When linearized, this becomes a system of 3 linear equations in 3 unknowns — exactly what we study here!
Real-world applications:
- Navigation: GPS, aircraft positioning, autonomous vehicles
- Economics: Supply-demand equilibrium, input-output analysis
- Engineering: Circuit analysis, structural loads, fluid dynamics
- Machine learning: Regression, neural network training
- Computer graphics: 3D rendering, animation keyframing
Why this matters for JEE: Linear systems appear in 4-5 questions in JEE Main and are crucial for coordinate geometry, calculus optimization, and physics problems in JEE Advanced.
Prerequisites
Before diving into linear equations, you should be comfortable with:
- Matrix Basics — Matrix notation and types
- Matrix Algebra — Matrix multiplication
- Determinants — Computing determinants
- Adjoint and Inverse — Matrix inverse
System of Linear Equations
General Form
A system of $m$ linear equations in $n$ unknowns:
$$\begin{cases} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m \end{cases}$$Matrix Form
$$\boxed{AX = B}$$where:
$$A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \quad B = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix}$$Components:
- $A$: Coefficient matrix ($m \times n$)
- $X$: Variable column matrix ($n \times 1$)
- $B$: Constant column matrix ($m \times 1$)
Homogeneous vs Non-Homogeneous
Homogeneous system: $AX = O$ (all constants are zero)
$$\begin{cases} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0 \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0 \\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0 \end{cases}$$Always has trivial solution $x_1 = x_2 = x_3 = 0$
Non-homogeneous system: $AX = B$ where $B \neq O$
$$\begin{cases} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = b_1 \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = b_2 \\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = b_3 \end{cases}$$Solution Methods for Square Systems
We focus on square systems where $m = n$ (same number of equations as unknowns).
Method 1: Matrix Method (Using Inverse)
Concept: If $AX = B$ and $A^{-1}$ exists:
$$\boxed{X = A^{-1}B}$$Steps:
- Check if $|A| \neq 0$ (inverse exists)
- Find $A^{-1} = \frac{1}{|A|} \text{adj}(A)$
- Multiply: $X = A^{-1}B$
Applicability: Only when $|A| \neq 0$ (unique solution exists)
Example: 2×2 System
Solve:
$$\begin{cases} 2x + 3y = 8 \\ x + 4y = 9 \end{cases}$$Step 1: Write in matrix form
$$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 9 \end{bmatrix}$$ $$A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 9 \end{bmatrix}$$Step 2: Find determinant
$$|A| = 8 - 3 = 5 \neq 0$$✓ (unique solution exists)
Step 3: Find inverse
$$A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$$Step 4: Solve
$$X = A^{-1}B = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 8 \\ 9 \end{bmatrix}$$ $$= \frac{1}{5} \begin{bmatrix} 32 - 27 \\ -8 + 18 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 5 \\ 10 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$Answer: $x = 1, y = 2$
Example: 3×3 System
Solve:
$$\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 4 \end{cases}$$Matrix form:
$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \\ 4 \end{bmatrix}$$Find $|A|$:
$$|A| = 1(1-2) - 1(-2-1) + 1(4+1) = -1 + 3 + 5 = 7 \neq 0$$✓
Find inverse and solve:
$$X = A^{-1}B$$(calculation omitted for brevity)
Answer: $x = 1, y = 2, z = 3$
Method 2: Cramer’s Rule
Cramer’s Rule provides an explicit formula for each variable using determinants.
For system $AX = B$ where $|A| \neq 0$:
$$\boxed{x_i = \frac{D_i}{D}}$$where:
- $D = |A|$ (determinant of coefficient matrix)
- $D_i$ = determinant of matrix formed by replacing $i^{th}$ column of $A$ with $B$
For 2×2 System
$$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$ $$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$$ $$D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} \quad \text{(replace column 1 with constants)}$$ $$D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} \quad \text{(replace column 2 with constants)}$$ $$\boxed{x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}}$$For 3×3 System
$$\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}$$ $$D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$ $$D_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}, \quad D_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}, \quad D_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$$ $$\boxed{x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}}$$Step 1: Find $D$ = determinant of coefficient matrix
Step 2: For each variable:
- Replace that variable’s column with the constant column
- Find the determinant
Step 3: Variable = $\frac{\text{New determinant}}{D}$
Mnemonic: “Divide Determinants Directly”
For $x$: Replace x-column (column 1) For $y$: Replace y-column (column 2) For $z$: Replace z-column (column 3)
Example: Using Cramer’s Rule
Solve:
$$\begin{cases} 2x + 3y = 8 \\ x + 4y = 9 \end{cases}$$Step 1: Find $D$
$$D = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 8 - 3 = 5$$Step 2: Find $D_x$ (replace column 1 with constants)
$$D_x = \begin{vmatrix} 8 & 3 \\ 9 & 4 \end{vmatrix} = 32 - 27 = 5$$Step 3: Find $D_y$ (replace column 2 with constants)
$$D_y = \begin{vmatrix} 2 & 8 \\ 1 & 9 \end{vmatrix} = 18 - 8 = 10$$Step 4: Solve
$$x = \frac{D_x}{D} = \frac{5}{5} = 1$$ $$y = \frac{D_y}{D} = \frac{10}{5} = 2$$Answer: $x = 1, y = 2$
Consistency of Linear Systems
Definitions
Consistent system: Has at least one solution
- Can have unique solution or infinitely many solutions
Inconsistent system: Has no solution
- Represents parallel lines/planes that don’t intersect
Consistency Conditions (Square Systems)
For system $AX = B$ where $A$ is $n \times n$:
| Condition | Nature | Solutions |
|---|---|---|
| $\|A\| \neq 0$ | Consistent | Unique solution |
| $\|A\| = 0$ and $(\text{adj } A)B \neq O$ | Inconsistent | No solution |
| $\|A\| = 0$ and $(\text{adj } A)B = O$ | Consistent | Infinitely many solutions |
To check consistency:
Step 1: Calculate $|A|$
If $|A| \neq 0$: STOP → Unique solution exists (use Cramer’s or matrix method)
If $|A| = 0$: Calculate $(\text{adj } A) \cdot B$
- If $(\text{adj } A)B \neq O$: Inconsistent (no solution)
- If $(\text{adj } A)B = O$: Infinitely many solutions (system is dependent)
Time-saver: For JEE, often you only need to check $|A|$ to determine uniqueness!
Example: Checking Consistency
Determine the nature of solutions:
$$\begin{cases} x + y + z = 4 \\ 2x + y - z = 1 \\ x - y + 2z = 2 \end{cases}$$Step 1: Find $|A|$
$$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 1 & -1 & 2 \end{bmatrix}$$ $$|A| = 1(2-1) - 1(4+1) + 1(-2-1) = 1 - 5 - 3 = -7 \neq 0$$Conclusion: Unique solution exists ✓
Homogeneous Systems
For homogeneous system $AX = O$:
Trivial Solution
Always has the trivial solution: $X = O$ (all variables = 0)
Non-Trivial Solutions
Condition for non-trivial solutions:
$$\boxed{|A| = 0}$$If $|A| = 0$: Infinitely many non-trivial solutions exist
If $|A| \neq 0$: Only trivial solution exists
Example: Homogeneous System
Find condition for non-trivial solutions:
$$\begin{cases} x + 2y + 3z = 0 \\ 2x + 3y + kz = 0 \\ 3x + 4y + 5z = 0 \end{cases}$$For non-trivial solutions, we need $|A| = 0$:
$$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & k \\ 3 & 4 & 5 \end{vmatrix} = 0$$Expanding:
$$1(15-4k) - 2(10-3k) + 3(8-9) = 0$$ $$15 - 4k - 20 + 6k - 3 = 0$$ $$2k - 8 = 0$$ $$k = 4$$Answer: For $k = 4$, the system has infinitely many non-trivial solutions.
Comparison of Methods
| Method | Pros | Cons | Best For |
|---|---|---|---|
| Matrix Method | ✓ General, systematic ✓ Good for symbolic | ✗ Need to find inverse ✗ Slow for large systems | Theoretical problems |
| Cramer’s Rule | ✓ Direct formulas ✓ Good for single variable | ✗ Many determinants ✗ Inefficient for $n > 3$ | Finding one variable |
| Elimination | ✓ Fast for numerical ✓ Works for rectangular | ✗ Not compact formula | Numerical computation |
For 2×2 systems:
- Use Cramer’s rule (fastest — 30 seconds)
For 3×3 systems:
- If finding all variables: Use matrix method or Cramer’s
- If finding one variable: Use Cramer’s rule for that variable
- If checking consistency: Find $|A|$ first
For theoretical problems:
- Use matrix method to show all steps clearly
For multiple choice:
- Sometimes you can substitute options to check!
Common Mistakes to Avoid
Wrong: Applying Cramer’s rule when $|A| = 0$
Right: Cramer’s rule only works when $|A| \neq 0$!
If $|A| = 0$, system is either inconsistent or has infinitely many solutions.
Wrong: For $D_y$, replacing row 2 with constants
Right: For $D_y$, replace column 2 with constants (not row!)
Variables correspond to columns, not rows.
Wrong: Homogeneous system has no solution when $|A| = 0$
Right: Homogeneous system always has at least the trivial solution $X = O$
When $|A| = 0$, it has infinitely many (non-trivial) solutions.
Wrong: If $|A| = 0$, system is always inconsistent
Right: If $|A| = 0$, check $(\text{adj } A)B$:
- If $\neq O$ → Inconsistent
- If $= O$ → Infinitely many solutions
Wrong: $X = BA^{-1}$ (wrong order)
Right: $X = A^{-1}B$ (inverse on left!)
Matrix multiplication is not commutative!
Practice Problems
Level 1: Foundation (NCERT)
Solve using matrix method:
$$\begin{cases} x + 2y = 5 \\ 2x + 3y = 8 \end{cases}$$Solution:
$$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \quad |A| = 3 - 4 = -1$$ $$A^{-1} = \frac{1}{-1} \begin{bmatrix} 3 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 2 \\ 2 & -1 \end{bmatrix}$$ $$X = A^{-1}B = \begin{bmatrix} -3 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$Answer: $x = 1, y = 2$
Solve using Cramer’s rule:
$$\begin{cases} 3x + 4y = 10 \\ 2x - 2y = 2 \end{cases}$$Solution:
$$D = \begin{vmatrix} 3 & 4 \\ 2 & -2 \end{vmatrix} = -6 - 8 = -14$$ $$D_x = \begin{vmatrix} 10 & 4 \\ 2 & -2 \end{vmatrix} = -20 - 8 = -28$$ $$D_y = \begin{vmatrix} 3 & 10 \\ 2 & 2 \end{vmatrix} = 6 - 20 = -14$$ $$x = \frac{-28}{-14} = 2, \quad y = \frac{-14}{-14} = 1$$Answer: $x = 2, y = 1$
Check if the system is consistent:
$$\begin{cases} x + y = 3 \\ 2x + 2y = 6 \end{cases}$$Solution:
$$|A| = \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = 2 - 2 = 0$$Row 2 = 2 × Row 1, so equations are dependent.
Answer: Consistent with infinitely many solutions (e.g., $x = t, y = 3-t$)
Level 2: JEE Main
Solve:
$$\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}$$Solution:
Using Cramer’s rule:
$$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} = 1(1-1) - 1(-1-2) + 1(1+2) = 0 + 3 + 3 = 6$$ $$D_x = \begin{vmatrix} 6 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 6(1-1) - 1(-2-1) + 1(2+1) = 0 + 3 + 3 = 6$$ $$x = \frac{6}{6} = 1$$Similarly: $y = 2, z = 3$
Answer: $x = 1, y = 2, z = 3$
For what value of $k$ does the system have no solution?
$$\begin{cases} x + 2y = 3 \\ 2x + ky = 5 \end{cases}$$Solution:
For no solution: $|A| = 0$ and system inconsistent
$$|A| = \begin{vmatrix} 1 & 2 \\ 2 & k \end{vmatrix} = k - 4 = 0 \Rightarrow k = 4$$Check: When $k = 4$, system becomes:
$$x + 2y = 3$$ $$2x + 4y = 5$$(which is $2(x+2y) = 2(3) = 6 \neq 5$)
Inconsistent! ✓
Answer: $k = 4$
Find condition for non-trivial solution:
$$\begin{cases} x + ky + 3z = 0 \\ 2x + 3y - z = 0 \\ 3x + 2y - 2z = 0 \end{cases}$$Solution:
For non-trivial solutions: $|A| = 0$
$$\begin{vmatrix} 1 & k & 3 \\ 2 & 3 & -1 \\ 3 & 2 & -2 \end{vmatrix} = 0$$ $$1(-6+2) - k(-4+3) + 3(4-9) = 0$$ $$-4 + k - 15 = 0$$ $$k = 19$$Answer: $k = 19$
Level 3: JEE Advanced
If the system has infinitely many solutions, find the relation between $a, b, c$:
$$\begin{cases} x + 2y + 3z = a \\ 2x + 3y + z = b \\ 3x + 4y + 2z = c \end{cases}$$Solution:
For infinitely many solutions: $|A| = 0$ and $(\text{adj } A)B = O$
$$|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 4 & 2 \end{vmatrix}$$ $$= 1(6-4) - 2(4-3) + 3(8-9) = 2 - 2 - 3 = -3 \neq 0$$Wait, this means unique solution exists for general $a, b, c$.
For infinitely many solutions, we need the augmented matrix to have rank defect.
Actually, the system has unique solution for all $a, b, c$ since $|A| \neq 0$.
Answer: No relation needed; system always has unique solution (question premise incorrect)
Solve the system if $\lambda = 2$:
$$\begin{cases} x + y + z = 1 \\ 2x + 3y + 2z = \lambda \\ x + 2y + 3z = 3 \end{cases}$$Solution:
Substitute $\lambda = 2$:
$$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$ $$|A| = 1(9-4) - 1(6-2) + 1(4-3) = 5 - 4 + 1 = 2 \neq 0$$Using matrix method or Cramer’s rule:
$$x = 0, \quad y = -1, \quad z = 2$$Answer: $x = 0, y = -1, z = 2$
Quick Revision Box
| Concept | Formula/Condition |
|---|---|
| Matrix form | $AX = B$ |
| Matrix method | $X = A^{-1}B$ (if $\|A\| \neq 0$) |
| Cramer’s rule | $x_i = \frac{D_i}{D}$ where $D = \|A\|$ |
| Unique solution | $\|A\| \neq 0$ |
| No solution | $\|A\| = 0$ and $(\text{adj } A)B \neq O$ |
| Infinite solutions | $\|A\| = 0$ and $(\text{adj } A)B = O$ |
| Homogeneous trivial | $X = O$ (always exists) |
| Homogeneous non-trivial | $\|A\| = 0$ (infinitely many) |
| Consistency check | Calculate $\|A\|$ first! |
Related Topics
Within Matrices & Determinants Chapter
- Matrix Basics — Matrix notation
- Matrix Algebra — Matrix operations
- Determinants — Computing $|A|$
- Adjoint and Inverse — Finding $A^{-1}$
- Properties of Determinants — Shortcuts for evaluation
Math Connections
- Coordinate Geometry — Intersection of lines
- Vector Algebra — Vector equations
- Complex Numbers — Complex linear systems
Real-World Applications
- GPS positioning — Satellite triangulation
- Economics — Input-output analysis, equilibrium
- Circuit analysis — Kirchhoff’s laws
- Structural engineering — Force distribution
- Machine learning — Linear regression
Teacher’s Summary
- Matrix form: $AX = B$ (compact representation)
- Two main methods: Matrix method ($X = A^{-1}B$) and Cramer’s rule ($x_i = \frac{D_i}{D}$)
- Consistency conditions:
- $|A| \neq 0$ → Unique solution
- $|A| = 0, (\text{adj } A)B \neq O$ → No solution
- $|A| = 0, (\text{adj } A)B = O$ → Infinitely many solutions
- Homogeneous systems: Always have trivial solution; non-trivial exists iff $|A| = 0$
- Cramer’s rule: Replace column with constants, divide by $D$
- Quick check: Always find $|A|$ first to determine nature of solutions
“First check the determinant — it tells you everything about the system!”
Exam Strategy:
- Step 1: Write in matrix form $AX = B$
- Step 2: Calculate $|A|$
- If $|A| \neq 0$ → Use Cramer’s or matrix method
- If $|A| = 0$ → Check consistency
- Step 3: For 2×2, Cramer’s is fastest (30 seconds)
- Step 4: For 3×3, choose based on what’s asked