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:
Add the first and second equations:
\[ (x + y + z) + (2x - y + 3z) = 6 + 14 \Rightarrow 3x + 4z = 20 \quad (A) \]
From the third equation, isolate \( z \): \( z = x + 4y - 2 \). Substitute this into the first equation:
\[ x + y + (x + 4y - 2) = 6 \Rightarrow 2x + 5y = 8 \quad (B) \]
Now, let's use elimination with equations (A) and (B) after expressing one variable in terms of others from (B):
From (B): \( y = \frac{8 - 2x}{5} \)
Substitute \( y \) and \( z = 2 - x + y \) into the original equations or use the derived equations (A) and (B).
Alternatively, from \( x + y + z = 6 \) (1) and \( x + 4y - z = 2 \) (3), add them:
\[ (x + y + z) + (x + 4y - z) = 6 + 2 \Rightarrow 2x + 5y = 8 \quad (B) \]
From \( x + y + z = 6 \) (1) and \( 2x - y + 3z = 14 \) (2), add them:
\[ (x + y + z) + (2x - y + 3z) = 6 + 14 \Rightarrow 3x + 4z = 20 \quad (A) \]
Let's use a simpler path to solve: Add (1) and (3): \( (x+y+z) + (x+4y-z) = 6+2 \Rightarrow 2x + 5y = 8 \quad (Eq. 4) \) Multiply (1) by 2: \( 2x + 2y + 2z = 12 \) Subtract (2) from this: \( (2x + 2y + 2z) - (2x - y + 3z) = 12 - 14 \Rightarrow 3y - z = -2 \quad (Eq. 5) \) From Eq. 5: \( z = 3y + 2 \). Substitute \( z \) into Eq. 1: \( x + y + (3y + 2) = 6 \Rightarrow x + 4y = 4 \quad (Eq. 6) \) Now we have a system for \( x \) and \( y \): \( 2x + 5y = 8 \quad (Eq. 4) \) \( x + 4y = 4 \quad (Eq. 6) \) From Eq. 6: \( x = 4 - 4y \). Substitute into Eq. 4: \( 2(4 - 4y) + 5y = 8 \Rightarrow 8 - 8y + 5y = 8 \Rightarrow -3y = 0 \Rightarrow y = 0 \). Then \( x = 4 - 4(0) = 4 \). Finally, \( z = 3y + 2 = 3(0) + 2 = 2 \).
Final solution: \( \boxed{x = 4, y = 0, z = 2} \)
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:
Using elimination or matrix method:
(1) \( 3x + 2y - z = 1 \)
(2) \( x - y + 2z = 3 \)
(3) \( 2x + y + z = 4 \)
Add (2) and (3): \( (x - y + 2z) + (2x + y + z) = 3 + 4 \Rightarrow 3x + 3z = 7 \quad (A) \)
From (3), \( y = 4 - 2x - z \). Substitute into (1):
\[ 3x + 2(4 - 2x - z) - z = 1 \] \[ 3x + 8 - 4x - 2z - z = 1 \] \[ -x - 3z = -7 \Rightarrow x + 3z = 7 \quad (B) \]
Now solve the system for \( x \) and \( z \):
\[ 3x + 3z = 7 \quad (A) \] \[ x + 3z = 7 \quad (B) \]
Subtract (B) from (A): \( (3x + 3z) - (x + 3z) = 7 - 7 \Rightarrow 2x = 0 \Rightarrow x = 0 \).
Substitute \( x = 0 \) into (B): \( 0 + 3z = 7 \Rightarrow z = \frac{7}{3} \).
Substitute \( x = 0 \) and \( z = \frac{7}{3} \) into (3):
\[ 2(0) + y + \frac{7}{3} = 4 \] \[ y = 4 - \frac{7}{3} = \frac{12 - 7}{3} = \frac{5}{3} \]
Final solution: \( \boxed{x = 0, y = \frac{5}{3}, z = \frac{7}{3}} \)
Solve the following system of linear equations:
Using elimination or matrix method:
(1) \( x + 2y + 3z = 10 \)
(2) \( 2x + 3y + z = 8 \)
(3) \( 3x + y + z = 7 \)
Subtract (2) from (3): \( (3x + y + z) - (2x + 3y + z) = 7 - 8 \Rightarrow x - 2y = -1 \quad (A) \)
Multiply (2) by 3: \( 6x + 9y + 3z = 24 \). Subtract (1) from this:
\[ (6x + 9y + 3z) - (x + 2y + 3z) = 24 - 10 \] \[ 5x + 7y = 14 \quad (B) \]
From (A): \( x = 2y - 1 \). Substitute into (B):
\[ 5(2y - 1) + 7y = 14 \] \[ 10y - 5 + 7y = 14 \] \[ 17y = 19 \Rightarrow y = \frac{19}{17} \]
Substitute \( y = \frac{19}{17} \) into \( x = 2y - 1 \):
\[ x = 2\left(\frac{19}{17}\right) - 1 = \frac{38}{17} - \frac{17}{17} = \frac{21}{17} \]
Substitute \( x = \frac{21}{17} \) and \( y = \frac{19}{17} \) into (3):
\[ 3\left(\frac{21}{17}\right) + \frac{19}{17} + z = 7 \] \[ \frac{63}{17} + \frac{19}{17} + z = 7 \] \[ \frac{82}{17} + z = 7 \] \[ z = 7 - \frac{82}{17} = \frac{119 - 82}{17} = \frac{37}{17} \]
Final solution: \( \boxed{x = \frac{21}{17}, y = \frac{19}{17}, z = \frac{37}{17}} \)
Solve the following system of linear equations:
From (1): \( x = y + 1 \), from (2): \( y = z + 2 \)
Then: \( x = (z + 2) + 1 \Rightarrow x = z + 3 \). This is identical to the third equation.
This system has infinitely many solutions. We can express them in terms of a parameter, say \( t \).
Let \( z = t \).
From \( y - z = 2 \Rightarrow y - t = 2 \Rightarrow y = t + 2 \).
From \( x - y = 1 \Rightarrow x - (t + 2) = 1 \Rightarrow x - t - 2 = 1 \Rightarrow x = t + 3 \).
Parametric solution: \( \boxed{x = t + 3,\ y = t + 2,\ z = t} \) for any real number \( t \).
Solve the following system of linear equations:
All equations are scalar multiples of the first equation \( \Rightarrow \) Infinitely many solutions.
Solve the following system of linear equations:
Using elimination or matrix method:
(1) \( 2x + 3y + z = 9 \)
(2) \( x - y + 2z = 8 \)
(3) \( 3x + 2y - z = 3 \)
Add (1) and (3): \( (2x + 3y + z) + (3x + 2y - z) = 9 + 3 \Rightarrow 5x + 5y = 12 \quad (A) \)
Multiply (3) by 2: \( 6x + 4y - 2z = 6 \). Add this to (2):
\[ (x - y + 2z) + (6x + 4y - 2z) = 8 + 6 \] \[ 7x + 3y = 14 \quad (B) \]
From (A): \( y = \frac{12 - 5x}{5} \). Substitute into (B):
\[ 7x + 3\left(\frac{12 - 5x}{5}\right) = 14 \] Multiply by 5 to clear denominator: \[ 35x + 3(12 - 5x) = 70 \] \[ 35x + 36 - 15x = 70 \] \[ 20x = 34 \Rightarrow x = \frac{34}{20} = \frac{17}{10} \]
Substitute \( x = \frac{17}{10} \) into \( y = \frac{12 - 5x}{5} \):
\[ y = \frac{12 - 5\left(\frac{17}{10}\right)}{5} = \frac{12 - \frac{17}{2}}{5} = \frac{\frac{24 - 17}{2}}{5} = \frac{\frac{7}{2}}{5} = \frac{7}{10} \]
Substitute \( x = \frac{17}{10} \) and \( y = \frac{7}{10} \) into (1):
\[ 2\left(\frac{17}{10}\right) + 3\left(\frac{7}{10}\right) + z = 9 \] \[ \frac{34}{10} + \frac{21}{10} + z = 9 \] \[ \frac{55}{10} + z = 9 \] \[ z = 9 - \frac{55}{10} = \frac{90 - 55}{10} = \frac{35}{10} = \frac{7}{2} \]
Final solution: \( \boxed{x = \frac{17}{10}, y = \frac{7}{10}, z = \frac{7}{2}} \)
Solve the following system of linear equations:
Using elimination or matrix method:
(1) \( x + y + z = 0 \)
(2) \( x - y + z = 2 \)
(3) \( 2x + 3y + z = 1 \)
Add (1) and (2): \( (x + y + z) + (x - y + z) = 0 + 2 \Rightarrow 2x + 2z = 2 \Rightarrow x + z = 1 \quad (A) \)
Subtract (2) from (1): \( (x + y + z) - (x - y + z) = 0 - 2 \Rightarrow 2y = -2 \Rightarrow y = -1 \).
Substitute \( y = -1 \) into (3):
\[ 2x + 3(-1) + z = 1 \] \[ 2x - 3 + z = 1 \] \[ 2x + z = 4 \quad (B) \]
Now solve the system for \( x \) and \( z \):
\[ x + z = 1 \quad (A) \] \[ 2x + z = 4 \quad (B) \]
Subtract (A) from (B): \( (2x + z) - (x + z) = 4 - 1 \Rightarrow x = 3 \).
Substitute \( x = 3 \) into (A): \( 3 + z = 1 \Rightarrow z = -2 \).
Final solution: \( \boxed{x = 3, y = -1, z = -2} \)
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:
Using elimination or matrix method:
(1) \( x + y + z = 3 \)
(2) \( x - y + z = 1 \)
(3) \( x + 2y - z = 4 \)
Subtract (2) from (1): \( (x + y + z) - (x - y + z) = 3 - 1 \Rightarrow 2y = 2 \Rightarrow y = 1 \).
Substitute \( y = 1 \) into (1): \( x + 1 + z = 3 \Rightarrow x + z = 2 \quad (A) \)
Substitute \( y = 1 \) into (3): \( x + 2(1) - z = 4 \Rightarrow x + 2 - z = 4 \Rightarrow x - z = 2 \quad (B) \)
Now solve the system for \( x \) and \( z \):
\[ x + z = 2 \quad (A) \] \[ x - z = 2 \quad (B) \]
Add (A) and (B): \( (x + z) + (x - z) = 2 + 2 \Rightarrow 2x = 4 \Rightarrow x = 2 \).
Substitute \( x = 2 \) into (A): \( 2 + z = 2 \Rightarrow z = 0 \).
Final solution: \( \boxed{x = 2, y = 1, z = 0} \)
Solve the following system of linear equations:
Second is a multiple of the first: Infinite solutions.
Solve the following system of linear equations:
Using elimination or matrix method:
(1) \( x - 2y + 3z = 9 \)
(2) \( 2x + y - z = 4 \)
(3) \( -x + y + z = -2 \)
Add (2) and (3): \( (2x + y - z) + (-x + y + z) = 4 + (-2) \Rightarrow x + 2y = 2 \quad (A) \)
Multiply (2) by 3: \( 6x + 3y - 3z = 12 \). Add this to (1):
\[ (x - 2y + 3z) + (6x + 3y - 3z) = 9 + 12 \] \[ 7x + y = 21 \quad (B) \]
From (B): \( y = 21 - 7x \). Substitute into (A):
\[ x + 2(21 - 7x) = 2 \] \[ x + 42 - 14x = 2 \] \[ -13x = -40 \Rightarrow x = \frac{40}{13} \]
Substitute \( x = \frac{40}{13} \) into \( y = 21 - 7x \):
\[ y = 21 - 7\left(\frac{40}{13}\right) = \frac{273 - 280}{13} = -\frac{7}{13} \]
Substitute \( x = \frac{40}{13} \) and \( y = -\frac{7}{13} \) into (3):
\[ -\frac{40}{13} - \frac{7}{13} + z = -2 \] \[ -\frac{47}{13} + z = -2 \] \[ z = -2 + \frac{47}{13} = \frac{-26 + 47}{13} = \frac{21}{13} \]
Final solution: \( \boxed{x = \frac{40}{13}, y = -\frac{7}{13}, z = \frac{21}{13}} \)
Solve the following system of linear equations:
Observe the first two equations:
(1) \( x + 2y + z = 6 \)
(2) \( 2x + 4y + 2z = 12 \)
Equation (2) is simply 2 times Equation (1) (\( 2(x + 2y + z) = 2(6) \Rightarrow 2x + 4y + 2z = 12 \)). This means the first two equations are dependent and represent the same plane.
Now consider the system formed by equation (1) and (3):
(1) \( x + 2y + z = 6 \)
(3) \( x + y + z = 5 \)
Subtract (3) from (1): \( (x + 2y + z) - (x + y + z) = 6 - 5 \Rightarrow y = 1 \).
Substitute \( y = 1 \) back into (1) and (3):
From (1): \( x + 2(1) + z = 6 \Rightarrow x + z = 4 \).
From (3): \( x + 1 + z = 5 \Rightarrow x + z = 4 \).
Since \( x + z = 4 \) holds for both equations (after substituting \( y=1 \)), the system is consistent and has infinitely many solutions. We can express them in terms of a parameter, say \( t \).
Let \( z = t \).
From \( x + z = 4 \Rightarrow x + t = 4 \Rightarrow x = 4 - t \).
The solution is \( \boxed{x = 4 - t,\ y = 1,\ z = t} \) for any real number \( t \).
Solve the following system of linear equations:
Subtract (1) from (2): \( (2x + y) - (x + y) = 3 - 1 \Rightarrow x = 2 \), then \( y = -1 \). Plug into (3): \( 2 + 2(-1) = 0 \). Since \( 0 \neq 4 \Rightarrow \) No solution.
Solve the following system of linear equations:
Using elimination or matrix method:
(1) \( 3x + 4y + 2z = 18 \)
(2) \( x - y + z = 2 \)
(3) \( 2x + y - z = 5 \)
Add (2) and (3): \( (x - y + z) + (2x + y - z) = 2 + 5 \Rightarrow 3x = 7 \Rightarrow x = \frac{7}{3} \).
From (2), \( z = 2 - x + y \). Substitute into (1):
\[ 3x + 4y + 2(2 - x + y) = 18 \] \[ 3x + 4y + 4 - 2x + 2y = 18 \] \[ x + 6y = 14 \]
Substitute \( x = \frac{7}{3} \) into \( x + 6y = 14 \):
\[ \frac{7}{3} + 6y = 14 \] \[ 6y = 14 - \frac{7}{3} = \frac{42 - 7}{3} = \frac{35}{3} \] \[ y = \frac{35}{18} \]
Substitute \( x = \frac{7}{3} \) and \( y = \frac{35}{18} \) into \( z = 2 - x + y \):
\[ z = 2 - \frac{7}{3} + \frac{35}{18} = \frac{36}{18} - \frac{42}{18} + \frac{35}{18} = \frac{36 - 42 + 35}{18} = \frac{29}{18} \]
Final solution: \( \boxed{x = \frac{7}{3}, y = \frac{35}{18}, z = \frac{29}{18}} \)
Solve the following system of linear equations:
Add equations: \[ (2x - y) + (3x + y) = 1 + 9 \Rightarrow 5x = 10 \Rightarrow x = 2 \Rightarrow y = 9 - 3x = 9 - 6 = 3 \]