1.10 Systems of Linear EquationsHL Only

1. Introduction

This section studies systems of equations, such as $2 \times 2$ systems ($a_1x + b_1y = c_1$) and $3 \times 3$ 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 the simple equation $ax = b$:

  • If $a \ne 0$: There is a unique solution $x = \dfrac{b}{a}$.
  • If $a = 0$ and $b \ne 0$: There is no solution (e.g., $0x = 5$).
  • If $a = 0$ and $b = 0$: There are infinitely many solutions ($0x = 0$ is true for any $x$).

2. 2x2 Systems

EXAMPLE 1

  • (a) Unique Solution:

    $$\begin{aligned} \begin{cases} 2x + 3y = 9 \\ 4x + 7y = 19 \end{cases} \implies (x,y) = (3,1) \end{aligned}$$

  • (b) No Solution (Parallel Lines):

    $$\begin{aligned} \begin{cases} x + 2y = 1 \\ 2x + 4y = 10 \end{cases} \implies x + 2y = 5 \end{aligned}$$

    This system is inconsistent because $1 \ne 5$.

  • (c) Infinitely Many Solutions (Coincident Lines):

    $$\begin{aligned} \begin{cases} x + 2y = 1 \\ 2x + 4y = 2 \end{cases} \implies x + 2y = 1 \end{aligned}$$

    The system reduces to exactly one equation. We can introduce a parameter: if $y = \lambda$, then $x = 1 - 2\lambda$.

    Solution set: $(1-2\lambda, \lambda)$, where $\lambda \in \mathbb{R}$.

3. 3x3 Systems

Gaussian Elimination can be used to solve these larger systems.

EXAMPLE 2

  • (a) Unique Solution:

    $$\begin{aligned} \begin{cases} 5x + 11y - 21z = -22 \\ x + 2y - 4z = -4 \\ 3x - 2y + 3z = 11 \end{cases} \end{aligned}$$

    Solving the system gives the unique coordinate: $(2, -1, 1)$.

  • (b) No Solution:

    $$\begin{aligned} \begin{cases} 2x + 3y + 3z = 3 \\ x + y - 2z = 4 \\ 5x + 7y + 4z = 5 \end{cases} \end{aligned}$$

    Row reduction results in an impossible equation (e.g., $0 = d$ where $d \ne 0$), indicating there is no solution.

  • (c) Infinitely Many Solutions:

    $$\begin{aligned} \begin{cases} 2x + 3y + 3z = 3 \\ x + y - 2z = 4 \\ 5x + 7y + 4z = 10 \end{cases} \end{aligned}$$

    Row reduction yields parameterized results because one equation is redundant. We set $z = \lambda$ (the free variable), yielding:

    $$x = 9 + 9\lambda, \quad y = -5 - 7\lambda, \quad z = \lambda$$

4. Gaussian Elimination

This method uses an augmented matrix to eliminate variables systematically without writing out the $x, y, z$ repeatedly.

Methodology:

  1. Write the augmented matrix of coefficients.
  2. Use row operations to create zeros below the main diagonal (upper triangular form).
    • Interchange rows ($R_1 \leftrightarrow R_2$)
    • Multiply a row by a non-zero scalar ($R_1 \to 5R_1$)
    • Add a multiple of one row to another ($R_2 \to R_2 - 5R_1$)
  3. Analyze the final row to determine the nature of the solution, and use back-substitution.

REMARK: Ideally, during step 2, we attempt to create matrices where the leading coefficient of the selected row is $1$ to make row operations easier.

EXAMPLE: Solving System (a) Manually

Original System:

$$\begin{aligned} \begin{cases} 5x + 11y - 21z = -22 \\ x + 2y - 4z = -4 \\ 3x - 2y + 3z = 11 \end{cases} \end{aligned}$$

Step 1: Set up the augmented matrix. To get a leading $1$ at the top left, interchange $R_1$ and $R_2$.

$$\left[ \begin{array}{ccc|c} 1 & 2 & -4 & -4 \\ 5 & 11 & -21 & -22 \\ 3 & -2 & 3 & 11 \end{array} \right]$$

Step 2: Eliminate $x$ from $R_2$ ($R_2 \to R_2 - 5R_1$) and $R_3$ ($R_3 \to R_3 - 3R_1$).

$$\left[ \begin{array}{ccc|c} 1 & 2 & -4 & -4 \\ 0 & 1 & -1 & -2 \\ 0 & -8 & 15 & 23 \end{array} \right]$$

Step 3: Eliminate $y$ from $R_3$ ($R_3 \to R_3 + 8R_2$).

$$\left[ \begin{array}{ccc|c} 1 & 2 & -4 & -4 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 7 & 7 \end{array} \right]$$

Step 4: Back Substitution

$$\begin{aligned} &\text{From } R_3: \quad 7z = 7 \implies z = 1 \\ &\text{From } R_2: \quad y - z = -2 \implies y - 1 = -2 \implies y = -1 \\ &\text{From } R_1: \quad x + 2y - 4z = -4 \implies x - 2 - 4 = -4 \implies x = 2 \end{aligned}$$

Solution: $(2, -1, 1)$.

Interpreting the Final Matrix

  • Case 1 (Unique Solution): Main diagonal is non-zero (like the example above).
  • Case 2 (No Solution): Last row takes the form $\left[ \begin{array}{ccc|c} 0 & 0 & 0 & d \end{array} \right]$ where $d \ne 0$. This implies $0x + 0y + 0z = d$, which is mathematically impossible.
  • Case 3 (Infinitely Many Solutions): Last row takes the form $\left[ \begin{array}{ccc|c} 0 & 0 & 0 & 0 \end{array} \right]$. This implies $0 = 0$ (which is always true). We must 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 a sequence of Gaussian elimination row operations yields the following matrix:

$$\left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 5 & 6 \\ 0 & 0 & a & b \end{array} \right]$$

  • (1) Unique Solution:
    If $a \ne 0$, there is a unique solution. We calculate $z = \dfrac{b}{a}$, and then substitute back upwards to find exact numbers for $y$ and $x$.
  • (2) No Solution:
    If $a = 0$ and $b \ne 0$, the last row gives the impossible statement $0 = b$.
  • (3) Infinitely Many Solutions:
    If $a = 0$ and $b = 0$, the last row reads $0 = 0$. Set the free variable $z = \lambda$.

    $$\begin{aligned} \text{From } R_2: \quad & y + 5\lambda = 6 \implies y = 6 - 5\lambda \\ \text{From } R_1: \quad & x + 2(6 - 5\lambda) + 3\lambda = 4 \\ & x + 12 - 10\lambda + 3\lambda = 4 \implies x = -8 + 7\lambda \end{aligned}$$

    General solution: $(-8 + 7\lambda, \, 6 - 5\lambda, \, \lambda)$.

Geometrical Interpretation

  • 2x2 System: Represents the intersection of two lines in a plane.
  • 3x3 System: Represents the intersection of three planes in 3D space.
Unique Solution (x, y) No Solution (Parallel) Infinite (Coincident)