ECE35 - Summer 2025
A linear system of equations is a collection of two or more linear equations involving the same set of variables. Solving these systems is a fundamental skill in both mathematics and engineering, with applications in circuit analysis, control systems, and signal processing.
A linear equation in variables \( x_1, x_2, \dots, x_n \) is an equation of the form:
\[ a_1x_1 + a_2x_2 + \dots + a_nx_n = b \]
A system of linear equations is a collection of such equations involving the same variables:
\[ \begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m \end{cases} \]
Any linear system can be written compactly in matrix form:
\[ A\mathbf{x} = \mathbf{b} \]
Where:
\( A = \text{coefficient matrix},\quad \mathbf{x} = \text{unknown vector},\quad \mathbf{b} = \text{constant vector} \)
Used for small systems or by hand. Solve one equation for one variable and substitute into others to eliminate variables.
If \( A \) is a square matrix and \( \det(A) \neq 0 \), then:
\[ \mathbf{x} = A^{-1}\mathbf{b} \]
Solve the system of equations:
\[ \begin{cases} 2x + 3y = 12 \\ 4x - y = 5 \end{cases} \]
Method 1: Substitution
From the second equation:
\[ y = 4x - 5 \]
Substitute into the first:
\[ 2x + 3(4x - 5) = 12 \Rightarrow 2x + 12x - 15 = 12 \Rightarrow 14x = 27 \Rightarrow x = \frac{27}{14} \]
\[ y = 4\left(\frac{27}{14}\right) - 5 = \frac{108}{14} - \frac{70}{14} = \frac{38}{14} = \frac{19}{7} \]
Method 2: Matrix Inverse
Write the system as:
\[ \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 12 \\ 5 \end{bmatrix} \]
Compute inverse of the coefficient matrix:
\[ \det = (2)(-1) - (3)(4) = -2 - 12 = -14 \]
\[ A^{-1} = \frac{1}{-14} \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \frac{1}{14} \begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix} \]
Then:
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{14} \begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 12 \\ 5 \end{bmatrix} = \frac{1}{14} \begin{bmatrix} 12 + 15 \\ 48 - 10 \end{bmatrix} = \begin{bmatrix} \frac{27}{14} \\ \frac{19}{7} \end{bmatrix} \]
Solve the system:
\[ \begin{cases} x + 2y + z = 6 \\ 2x + 3y + 3z = 14 \\ x + y + z = 8 \end{cases} \]
Method 1: Elimination
From the first and third equations, subtract to eliminate \( x \):
\[ (x + 2y + z) - (x + y + z) = 6 - 8 \Rightarrow y = -2 \]
Substitute into the first:
\[ x + 2(-2) + z = 6 \Rightarrow x - 4 + z = 6 \Rightarrow x + z = 10 \quad (1) \]
Substitute \( y = -2 \) into the second:
\[ 2x + 3(-2) + 3z = 14 \Rightarrow 2x - 6 + 3z = 14 \Rightarrow 2x + 3z = 20 \quad (2) \]
From (1): \( x = 10 - z \). Substitute into (2):
\[ 2(10 - z) + 3z = 20 \Rightarrow 20 - 2z + 3z = 20 \Rightarrow z = 0 \Rightarrow x = 10 \]
\[ \boxed{x = 10,\ y = -2,\ z = 0} \]
Method 2: Matrix Inverse
\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 3 \\ 1 & 1 & 1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 6 \\ 14 \\ 8 \end{bmatrix} \]
Finally find:
\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} 10 \\ -2 \\ 0 \end{bmatrix} \]
Determine the type of solution for the system:
\[ \begin{cases} x + y = 4 \\ 2x + 2y = 8 \end{cases} \]
Method 1: Elimination
Multiply the first equation by 2:
\[ 2x + 2y = 8 \Rightarrow \text{same as second equation} \]
So, the system is dependent and has infinitely many solutions.
Method 2: Matrix Rank Check
\[ A = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix},\quad \mathbf{b} = \begin{bmatrix} 4 \\ 8 \end{bmatrix} \]
Rank of \( A \) = 1 < number of variables → Infinitely many solutions.
Solve the following system of linear equations:
From the first equation: \( x = 7 - 2y \)
Substitute into the second: \[ 3(7 - 2y) - y = 5 \Rightarrow 21 - 6y - y = 5 \Rightarrow -7y = -16 \Rightarrow y = \frac{16}{7} \]
Then: \[ x = 7 - 2\cdot\frac{16}{7} = \frac{49 - 32}{7} = \frac{17}{7} \]
Solve the following system of linear equations:
Note: Second equation is a multiple of the first \( (2x - 3y = 4) \Rightarrow \) Infinite solutions.
Solve the following system of linear equations:
Use substitution or matrix. Final solution: \( x = 3, y = 1, z = 2 \)
Solve the following system of linear equations:
Multiply the first equation by 2: \( 4x + 2y = 10 \Rightarrow 10 \neq 12 \Rightarrow \) No solution.
Solve the following system of linear equations:
Solve system using substitution or elimination. Final: \( x = 1, y = 2, z = -1 \)
Solve the following system of linear equations:
Matrix or elimination method yields: \( x = 1, y = 2, z = 1 \)
Solve the following system of linear equations:
From (1): \( x = y + 1 \), from (2): \( y = z + 2 \)
Then: \( x = z + 3 \). Parametric solution:
\[ x = z + 3,\quad y = z + 2 \]
Solve the following system of linear equations:
All equations are scalar multiples \( \Rightarrow \) Infinitely many solutions.
Solve the following system of linear equations:
Solving via substitution or matrix: \( x = 1, y = 2, z = 0 \)
Solve the following system of linear equations:
Subtract equations, solve system: \[ x = 1, y = -1, z = 0 \]
Solve the following system of linear equations:
From (1): \( x = 10 - 3y \)
Substitute into (2): \[ 2(10 - 3y) - y = 4 \Rightarrow 20 - 6y - y = 4 \Rightarrow 7y = 16 \Rightarrow y = \frac{16}{7}, \quad x = 10 - \frac{48}{7} = \frac{22}{7} \]
Solve the following system of linear equations:
From (2): \( x = -3 + 2y \), plug into (1): \[ 4(-3 + 2y) + y = 9 \Rightarrow -12 + 8y + y = 9 \Rightarrow 9y = 21 \Rightarrow y = \frac{7}{3}, x = -3 + \frac{14}{3} = \frac{5}{3} \]
Solve the following system of linear equations:
From (2): \( y = 7 - 3x \), plug into (1): \[ 5x - 2(7 - 3x) = 3 \Rightarrow 5x - 14 + 6x = 3 \Rightarrow 11x = 17 \Rightarrow x = \frac{17}{11}, y = 7 - \frac{51}{11} = \frac{26}{11} \]
Solve the following system of linear equations:
Subtract (1) and (2): \( 2y = 2 \Rightarrow y = 1 \)
Use in (1): \( x + 1 + z = 3 \Rightarrow x + z = 2 \)
Use in (2): \( x - 1 + z = 1 \Rightarrow x + z = 2 \)
Consistent, then find: \( x = 1, z = 1 \)
Solve the following system of linear equations:
Second is a multiple of the first: Infinite solutions.
Solve the following system of linear equations:
Solving yields: \( x = 2, y = 1, z = 1 \)
Solve the following system of linear equations:
First and second are multiples \( \Rightarrow \) consistent. Third not consistent \( \Rightarrow \) No solution.
Solve the following system of linear equations:
Subtract (1) from (2): \( x = 2 \), then \( y = -1 \). Plug into (3): \( 2 + 2(-1) \neq 4 \Rightarrow \) No solution.
Solve the following system of linear equations:
Use elimination or matrix method. Final: \( x = 2, y = 1, z = 1 \)
Solve the following system of linear equations:
Add equations: \[ (2x - y) + (3x + y) = 1 + 9 \Rightarrow 5x = 10 \Rightarrow x = 2 \Rightarrow y = 3x - 9 = -3 \]