System of Linear equations

Leave a Comment / By /

Back to: Mathematics – Linear Algebra

0
System of Linear Equations – Linear Algebra

System of Linear Equations

😊 Hello, students! Linear systems are the backbone of applied mathematics — from engineering to economics, from computer graphics to machine learning. Today, we’ll explore them deeply: how to represent them, solve them, and understand when solutions exist — and how many!

1. What Is a Linear Equation?

A linear equation in variables \( x_1, x_2, \dots, x_n \) is an equation of the form: \[ a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b \] where \( a_1, \dots, a_n, b \in \mathbb{R} \) are constants.

It has no products, powers, or nonlinear functions (like \( x^2 \), \( \sin x \), \( xy \)).

Linear: \( 2x – 3y + z = 5 \)

Not linear: \( x^2 + y = 1 \), \( xy = 2 \), \( \sin(x) + y = 0 \)

2. System of Linear Equations

A system of linear equations is a finite set of linear equations in the same variables. For example, a system of \( m \) equations in \( n \) unknowns: \[ \begin{aligned} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\ &\vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \end{aligned} \]

3. Matrix Representation

We can write the system compactly as:

\[ A\mathbf{x} = \mathbf{b} \]

where:

  • \( A = (a_{ij})_{m \times n} \) is the coefficient matrix
  • \( \mathbf{x} = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} \) is the unknown vector
  • \( \mathbf{b} = \begin{bmatrix} b_1 \\ \vdots \\ b_m \end{bmatrix} \) is the constant vector

System:

\[ \begin{cases} x + 2y – z = 4 \\ 3x – y + 2z = 1 \end{cases} \]

Matrix form:

\[ \begin{bmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix} \]

4. Augmented Matrix

The augmented matrix combines \( A \) and \( \mathbf{b} \): \[ [A \mid \mathbf{b}] \] It fully encodes the system.

For the system above:

\[ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 3 & -1 & 2 & 1 \end{array}\right] \]

5. Homogeneous vs. Non-Homogeneous

Homogeneous system: \( \mathbf{b} = \mathbf{0} \)

\[ A\mathbf{x} = \mathbf{0} \]

→ Always has at least the trivial solution \( \mathbf{x} = \mathbf{0} \).

Non-homogeneous system: \( \mathbf{b} \ne \mathbf{0} \)

→ May have no solution, one, or infinitely many.

6. Consistency and Number of Solutions

A system is:

  • Consistent if it has at least one solution.
  • Inconsistent if it has no solution.

If consistent, it has either:

  • Exactly one solution (unique), or
  • Infinitely many solutions.

🤔 Geometric view (in \( \mathbb{R}^2 \)):
  • Two lines intersect → unique solution
  • Parallel lines → no solution
  • Coincident lines → infinitely many solutions

7. Rank and Consistency

Let \( A \) be the coefficient matrix and \( [A \mid \mathbf{b}] \) the augmented matrix.

The system \( A\mathbf{x} = \mathbf{b} \) is consistent if and only if:

\[ \text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) \]

If consistent and \( A \) is \( m \times n \):

  • Unique solution ⇔ \( \text{rank}(A) = n \)
  • Infinitely many solutions ⇔ \( \text{rank}(A) < n \)

8. Solving Systems: Gaussian Elimination

Gaussian elimination uses elementary row operations to reduce the augmented matrix to row echelon form (REF) or reduced row echelon form (RREF).

Steps:

  1. Form the augmented matrix \( [A \mid \mathbf{b}] \)
  2. Use row ops to get REF
  3. Check for consistency (look for a row \( [0\ \cdots\ 0 \mid c] \) with \( c \ne 0 \))
  4. If consistent, use back-substitution (or go to RREF for direct read-off)

Example: Solve \[ \begin{cases} x + y + z = 6 \\ 2x + 3y + z = 10 \\ x + y + 2z = 8 \end{cases} \]

Augmented matrix:

\[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 3 & 1 & 10 \\ 1 & 1 & 2 & 8 \end{array}\right] \xrightarrow{R_2 – 2R_1,\ R_3 – R_1} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 1 & 2 \end{array}\right] \]

Back-substitute:

  • \( z = 2 \)
  • \( y – z = -2 \Rightarrow y = 0 \)
  • \( x + y + z = 6 \Rightarrow x = 4 \)

✅ Unique solution: \( (4, 0, 2) \)

😠 Red flag: If you get a row like \( [0\ 0\ 0 \mid 5] \), that means \( 0 = 5 \) → **inconsistent!**

9. Infinitely Many Solutions: Parametric Form

System: \[ \begin{cases} x + y = 2 \\ 2x + 2y = 4 \end{cases} \]

Augmented matrix → RREF:

\[ \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right] \]

Let \( y = t \) (free variable). Then \( x = 2 – t \).

General solution: \[ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} + t \begin{bmatrix} -1 \\ 1 \end{bmatrix}, \quad t \in \mathbb{R} \]

10. Summary Table

Condition Consistent? Solutions
\( \text{rank}(A) \ne \text{rank}([A \mid \mathbf{b}]) \) No None
\( \text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) = n \) Yes Unique
\( \text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) < n \) Yes Infinitely many

11. Self-Check: Test Your Understanding

  1. Can a homogeneous system be inconsistent?
  2. What does a row \( [0\ 0\ 0 \mid 0] \) mean in RREF?
  3. Solve: \[ \begin{cases} x + 2y = 5 \\ 2x + 4y = 10 \end{cases} \]

Answers: (1) No — always has \( \mathbf{0} \); (2) dependent equation; (3) infinitely many: \( x = 5 – 2t,\ y = t \)

System of Linear Equations – Questions & Full Solutions

System of Linear Equations – Questions & Full Solutions

😊 Hello, students! Below are all the system of linear equations questions from your documents, each with a complete, detailed solution using Gaussian elimination, RREF, and rank criteria. Use this to master consistency, uniqueness, and solution methods!

Q1. Determine if the following system is consistent. If so, how many solutions does it have?

\[ \begin{cases} x_1 – x_2 + x_3 = 1 \\ 3x_1 – 5x_2 + 5x_3 = 2 \\ 2x_1 – 2x_2 + 2x_3 = 5 \end{cases} \]

Solution:

Augmented matrix:

\[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 3 & -5 & 5 & 2 \\ 2 & -2 & 2 & 5 \end{array}\right] \]

Row reduce:

  • \( R_2 \to R_2 – 3R_1 \), \( R_3 \to R_3 – 2R_1 \)
\[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & -2 & 2 & -1 \\ 0 & 0 & 0 & 3 \end{array}\right] \]

Last row: \( 0 = 3 \) → **inconsistent**.

✅ No solution.

Q2. Solve the system:

\[ \begin{cases} x + 3y – z = 2 \\ 2x + 5y + 3z = 8 \\ -x – y = -1 \end{cases} \]

Solution:

Augmented matrix:

\[ \left[\begin{array}{ccc|c} 1 & 3 & -1 & 2 \\ 2 & 5 & 3 & 8 \\ -1 & -1 & 0 & -1 \end{array}\right] \]

Row reduce to RREF:

\[ \xrightarrow{\text{RREF}} \left[\begin{array}{ccc|c} 1 & 0 & -1 & -1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

Let \( z = t \) (free variable). Then:

  • \( y = 2 – t \)
  • \( x = -1 + t \)

✅ General solution: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}, \quad t \in \mathbb{R} \]

Q3. Solve:

\[ \begin{cases} 2x + y – z = 5 \\ x – y + 2z = 1 \\ 3x + 2y – z = 8 \end{cases} \]

Solution:

Augmented matrix → RREF:

\[ \xrightarrow{\text{RREF}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right] \]

✅ Unique solution: \( x = 2,\ y = 1,\ z = 0 \).

Q4. Express as \( A\mathbf{x} = \mathbf{b} \):

\[ \begin{cases} 5x + 4y – 3z = 7 \\ 4x + 3y – 2z = 5 \\ 3x + 2y – z = 5 \end{cases} \]

Solution:

\[ A = \begin{bmatrix} 5 & 4 & -3 \\ 4 & 3 & -2 \\ 3 & 2 & -1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 7 \\ 5 \\ 5 \end{bmatrix} \]

So the system is \( A\mathbf{x} = \mathbf{b} \).

Q5. Solve the homogeneous system:

\[ \begin{cases} x + y – z = 0 \\ 2x + 2y – 2z = 0 \\ 3x + 3y – 3z = 0 \end{cases} \]

Solution:

All equations are multiples of the first.

RREF of coefficient matrix:

\[ \begin{bmatrix} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

Let \( y = s,\ z = t \). Then \( x = -s + t \).

✅ General solution: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = s \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \quad s,t \in \mathbb{R} \]

Q6. For what value of \( \lambda \) is the system consistent?

\[ \begin{cases} x + y + z = 6 \\ 2x + 3y + z = 10 \\ x + y + \lambda z = 8 \end{cases} \]

Solution:

Augmented matrix:

\[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 3 & 1 & 10 \\ 1 & 1 & \lambda & 8 \end{array}\right] \xrightarrow{\text{row reduce}} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & \lambda – 1 & 2 \end{array}\right] \]

For consistency, we must not have \( 0 = \text{non-zero} \).

If \( \lambda = 1 \), last row becomes \( [0\ 0\ 0\ |\ 2] \) → inconsistent.

✅ So the system is consistent for all \( \lambda \ne 1 \).

(When \( \lambda \ne 1 \), unique solution exists.)

Q7. Solve:

\[ \begin{cases} x + 2y = 5 \\ 2x + 4y = 10 \end{cases} \]

Solution:

Second equation is \( 2 \times \) first → dependent.

RREF:

\[ \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 0 & 0 \end{array}\right] \]

Let \( y = t \), then \( x = 5 – 2t \).

✅ Solution: \( (x, y) = (5, 0) + t(-2, 1),\ t \in \mathbb{R} \).

Q8. Show the system has no solution:

\[ \begin{cases} x + y = 2 \\ 2x + 2y = 5 \end{cases} \]

Solution:

Augmented matrix:

\[ \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \xrightarrow{R_2 – 2R_1} \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right] \]

Last row: \( 0 = 1 \) → **inconsistent**.

✅ No solution.

📌 Key Takeaways:

  • Use RREF of augmented matrix to determine consistency.
  • System is consistent ⇔ no row of form \( [0\ \cdots\ 0\ |\ c],\ c \ne 0 \).
  • Free variables → infinitely many solutions.
  • Homogeneous systems always have at least the trivial solution.

Leave a Reply

Your email address will not be published. Required fields are marked *

Scroll to Top