1.10 Systems of Simultaneous Linear Equations
1. Introduction
This section studies systems of equations, such as 2x2 systems ($a_1x + b_1y = c_1$) and 3x3 systems ($a_1x + b_1y + c_1z = d_1$).
For any $n \times n$ system, there are exactly three possibilities for the number of solutions:
- A unique solution (Consistent)
- No solution (Inconsistent)
- Infinitely many solutions (Consistent)
Example (1x1 System): For $ax = b$:
- If $a \ne 0$: Unique solution $x = b/a$.
- If $a = 0$ and $b \ne 0$: No solution (e.g., $0x = 5$).
- If $a = 0$ and $b = 0$: Infinitely many solutions ($0x = 0$ is true for any $x$).
2. 2x2 Systems
EXAMPLE 1
$2x + 3y = 9$
$4x + 7y = 19$
Solution: $(x,y) = (3,1)$.
$x + 2y = 1$
$2x + 4y = 10 \Rightarrow x + 2y = 5$
Inconsistent ($1 \ne 5$).
$x + 2y = 1$
$2x + 4y = 2 \Rightarrow x + 2y = 1$
Reduces to one equation. If $y = \lambda$, then $x = 1 - 2\lambda$.
Solution set: $(1-2\lambda, \lambda), \lambda \in \mathbb{R}$.
3. 3x3 Systems
The GDC (Graphics Display Calculator) or Gaussian Elimination can be used to solve these systems.
EXAMPLE 2
$5x + 11y - 21z = -22$
$x + 2y - 4z = -4$
$3x - 2y + 3z = 11$
GDC gives: $(2, -1, 1)$.
$2x + 3y + 3z = 3$
$x + y - 2z = 4$
$5x + 7y + 4z = 5$
GDC indicates no solution.
$2x + 3y + 3z = 3$
$x + y - 2z = 4$
$5x + 7y + 4z = 10$
GDC gives: $x = 9 + 9\lambda, \quad y = -5 - 7\lambda, \quad z = \lambda$.
4. Gaussian Elimination
This method uses an augmented matrix to eliminate variables systematically.
Methodology:
- Write the augmented matrix of coefficients.
- Use row operations to create zeros below the main diagonal (triangular form).
- Interchange rows ($R_1 \leftrightarrow R_2$)
- Multiply a row by a scalar ($R_1 \to 5R_1$)
- Add a multiple of one row to another ($R_2 \to R_2 - 5R_1$)
- Analyze the final row to determine the nature of the solution.
Solving System (a) Manually
System:
$5x + 11y - 21z = -22$
$x + 2y - 4z = -4$
$3x - 2y + 3z = 11$
Step 1: Interchange R1 and R2 to get a leading 1.
[ 5 11 -21 | -22 ]
[ 3 -2 3 | 11 ]
Step 2: Eliminate $x$ from R2 ($R_2 - 5R_1$) and R3 ($R_3 - 3R_1$).
[ 0 1 -1 | -2 ]
[ 0 -8 15 | 23 ]
Step 3: Eliminate $y$ from R3 ($R_3 + 8R_2$).
[ 0 1 -1 | -2 ]
[ 0 0 7 | 7 ]
Step 4: Back Substitution
- From R3: $7z = 7 \Rightarrow z = 1$
- From R2: $y - z = -2 \Rightarrow y - 1 = -2 \Rightarrow y = -1$
- From R1: $x + 2y - 4z = -4 \Rightarrow x - 2 - 4 = -4 \Rightarrow x = 2$
Solution: $(2, -1, 1)$.
Interpreting the Final Matrix
Case 1 (Unique): Main diagonal is non-zero (like above).
Case 2 (No Solution): Last row is $[0 \quad 0 \quad 0 \quad | \quad d]$ with $d \ne 0$.
This implies $0x + 0y + 0z = d$, which is impossible.
Case 3 (Infinite Solutions): Last row is $[0 \quad 0 \quad 0 \quad | \quad 0]$.
This implies $0 = 0$ (always true). We assign a free variable (e.g., $z = \lambda$) and solve for $x$ and $y$ in terms of $\lambda$.
5. System with Parameters
EXAMPLE 3
Suppose Gaussian elimination yields the matrix:
[ 0 1 5 | 6 ]
[ 0 0 a | b ]
If $a \ne 0$, there is a unique solution.
$z = b/a$. Substitute back to find $y$ and $x$.
If $a = 0$ and $b \ne 0$, the last row gives $0 = b$, which is impossible.
If $a = 0$ and $b = 0$, the last row is $0 = 0$.
Set $z = \lambda$.
From R2: $y + 5\lambda = 6 \Rightarrow y = 6 - 5\lambda$.
From R1: $x + 2(6 - 5\lambda) + 3\lambda = 4 \Rightarrow x = -8 + 7\lambda$.
General solution: $(-8 + 7\lambda, 6 - 5\lambda, \lambda)$.
⚠️ Geometrical Interpretation
- 2x2 System: Represents the intersection of two lines in a plane. (Unique point, Parallel/No intersection, or Coincident/Infinite).
- 3x3 System: Represents the intersection of three planes in 3D space.