Systems of Linear Equations

Consider the system: \[ \begin{cases} 2x + 3y = 5 \\ 4x - y = 1 \end{cases} \] Matrix form: \[ A = \begin{bmatrix}2 & 3 \\ 4 & -1\end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix}x \\ y\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}5 \\ 1\end{bmatrix}. \]

System: \[ \begin{cases} x + 2y + z = 4 \\ 2x - y + 3z = 5 \\ -x + y + z = 1 \end{cases} \] Matrix form: \[ A = \begin{bmatrix}1 & 2 & 1 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix}x \\ y \\ z\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}4 \\ 5 \\ 1\end{bmatrix}. \]

System: \[ 3x - y + 2z = 7, \quad x + y - z = 0 \] Matrix form: \[ A = \begin{bmatrix}3 & -1 & 2 \\ 1 & 1 & -1\end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix}x \\ y \\ z\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}7 \\ 0\end{bmatrix}. \]

Homogeneous system: \[ x + y - z = 0, \quad 2x - y + 3z = 0 \]

Solution:

Form the augmented matrix (homogeneous system → last column is 0): \[ \begin{bmatrix} 1 & 1 & -1 & 0 \\ 2 & -1 & 3 & 0 \end{bmatrix} \]

Use row operations to reduce: \[ R_2 \to R_2 - 2R_1: \begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & -3 & 5 & 0 \end{bmatrix} \]

From second row: $-3y + 5z = 0 \implies y = \frac{5}{3} z$

From first row: $x + y - z = 0 \implies x + \frac{5}{3}z - z = 0 \implies x = -\frac{2}{3} z$

General solution: \[ \mathbf{x} = z \begin{bmatrix}-2/3 \\ 5/3 \\ 1\end{bmatrix}, \quad z \in \mathbb{R} \]

- Trivial solution: $x=y=z=0$ - Non-trivial solutions exist for $z \neq 0$.

Non-homogeneous system: \[ x + 2y = 3, \quad 3x - y = 4 \]

Solution:

Matrix form: $A \mathbf{x} = \mathbf{b}$, where \[ A = \begin{bmatrix}1 & 2 \\ 3 & -1\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}3 \\ 4\end{bmatrix} \]

Determinant: \[ \det(A) = 1\cdot(-1) - 3\cdot2 = -1 - 6 = -7 \neq 0 \]

Since $\det(A) \neq 0$, unique solution exists. Using Cramer's Rule:

\[ x = \frac{\det \begin{bmatrix}3 & 2 \\ 4 & -1\end{bmatrix}}{-7} = \frac{-3-8}{-7} = \frac{-11}{-7} = \frac{11}{7}, \quad y = \frac{\det \begin{bmatrix}1 & 3 \\ 3 & 4\end{bmatrix}}{-7} = \frac{4-9}{-7} = \frac{-5}{-7} = \frac{5}{7} \]

Solution: $x = 11/7$, $y = 5/7$

Homogeneous system with multiple solutions: \[ x + y + z = 0, \quad 2x + 3y + z = 0 \]

Solution:

Form augmented matrix: \[ \begin{bmatrix} 1 & 1 & 1 & 0 \\ 2 & 3 & 1 & 0 \end{bmatrix} \]

Row reduction: \[ R_2 \to R_2 - 2R_1: \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \end{bmatrix} \]

From second row: $y - z = 0 \implies y = z$

From first row: $x + y + z = 0 \implies x + z + z = 0 \implies x = -2z$

General solution: \[ \mathbf{x} = z \begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}, \quad z \in \mathbb{R} \]

- Rank$(A) = 2$, number of unknowns = 3 $\implies$ dimension of solution space = 1 - Infinitely many non-trivial solutions exist.

System: \[ x + y = 2, \quad 2x + 2y = 4 \]

Solution:

Matrix form: \[ A = \begin{bmatrix}1 & 1 \\ 2 & 2\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}2 \\ 4\end{bmatrix} \]

Augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix}1 & 1 & 2 \\ 2 & 2 & 4\end{bmatrix} \]

Row reduction: \[ R_2 \to R_2 - 2R_1 \implies \begin{bmatrix}1 & 1 & 2 \\ 0 & 0 & 0\end{bmatrix} \]

Ranks: \[ \text{Rank}(A) = 1, \quad \text{Rank}([A|\mathbf{b}]) = 1 \]

Since Rank$(A) = $ Rank$([A|\mathbf{b}]) < $ number of unknowns (2), system is consistent with infinitely many solutions.

General solution: Let $y = t$, then $x = 2 - t$, $t \in \mathbb{R}$.

System: \[ x + y = 2, \quad 2x + 2y = 5 \]

Solution:

Matrix form: \[ A = \begin{bmatrix}1 & 1 \\ 2 & 2\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}2 \\ 5\end{bmatrix} \]

Augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix}1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix} \]

Row reduction: \[ R_2 \to R_2 - 2R_1 \implies \begin{bmatrix}1 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix} \]

Ranks: \[ \text{Rank}(A) = 1, \quad \text{Rank}([A|\mathbf{b}]) = 2 \]

Since Rank$(A) \neq $ Rank$([A|\mathbf{b}])$, the system is inconsistent.

System: \[ x + y + z = 6, \quad x - y + z = 2, \quad 2x + z = 5 \]

Solution:

Augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 1 & -1 & 1 & 2 \\ 2 & 0 & 1 & 5 \end{bmatrix} \]

Step 1: $R_2 \to R_2 - R_1$, $R_3 \to R_3 - 2R_1$: \[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & -2 & 0 & -4 \\ 0 & -2 & -1 & -7 \end{bmatrix} \]

Step 2: $R_3 \to R_3 - R_2$: \[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & -2 & 0 & -4 \\ 0 & 0 & -1 & -3 \end{bmatrix} \]

Ranks: \[ \text{Rank}(A) = 3, \quad \text{Rank}([A|\mathbf{b}]) = 3 \]

Since Rank$(A) = $ Rank$([A|\mathbf{b}]) = $ number of unknowns, the system is consistent with a unique solution.

Step 3: Back-substitution: \[ z = 3, \quad -2y = -4 \implies y = 2, \quad x + y + z = 6 \implies x + 2 + 3 = 6 \implies x = 1 \]

Solution: $x=1$, $y=2$, $z=3$

Solve: \[ x + y + z = 6, \quad 2x + y + 3z = 10, \quad x + 2y + 3z = 11 \]

Solution:

Form augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 2 & 1 & 3 & 10 \\ 1 & 2 & 3 & 11 \end{bmatrix} \]

Step 1: $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$: \[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & -1 & 1 & -2 \\ 0 & 1 & 2 & 5 \end{bmatrix} \]

Step 2: $R_3 \to R_3 + R_2$: \[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & -1 & 1 & -2 \\ 0 & 0 & 3 & 3 \end{bmatrix} \]

Step 3: Back-substitution: \[ 3z = 3 \implies z = 1 \] \[ -R_2: -y + z = -2 \implies -y + 1 = -2 \implies y = 3 \] \[ R_1: x + y + z = 6 \implies x + 3 + 1 = 6 \implies x = 2 \]

Solution: $x=2$, $y=3$, $z=1$

System: \[ x + y = 3, \quad 2x + 3y = 7 \]

Solution:

Step 1: Eliminate $x$ from second equation: \[ 2x + 3y - 2(x + y) = 7 - 6 \implies y = 1 \]

Step 2: Substitute $y = 1$ into first equation: \[ x + 1 = 3 \implies x = 2 \]

Solution: $x=2$, $y=1$

System: \[ x + 2y + z = 5, \quad 3x + y + 2z = 7, \quad 2x + 3y + 3z = 10 \]

Solution:

Augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix} 1 & 2 & 1 & 5 \\ 3 & 1 & 2 & 7 \\ 2 & 3 & 3 & 10 \end{bmatrix} \]

Step 1: $R_2 \to R_2 - 3R_1$, $R_3 \to R_3 - 2R_1$: \[ \begin{bmatrix} 1 & 2 & 1 & 5 \\ 0 & -5 & -1 & -8 \\ 0 & -1 & 1 & 0 \end{bmatrix} \]

Step 2: $R_3 \to R_3 - \frac{-1}{-5}R_2 = R_3 - \frac{1}{5}R_2$: \[ \begin{bmatrix} 1 & 2 & 1 & 5 \\ 0 & -5 & -1 & -8 \\ 0 & 0 & 6/5 & 8/5 \end{bmatrix} \]

Step 3: Back-substitution: \[ \frac{6}{5} z = \frac{8}{5} \implies z = \frac{4}{3} \] \[ -5y - z = -8 \implies -5y - \frac{4}{3} = -8 \implies y = \frac{4}{3} \] \[ x + 2y + z = 5 \implies x + 2(\frac{4}{3}) + \frac{4}{3} = 5 \implies x = \frac{1}{3} \]

Solution: $x = \frac{1}{3}$, $y = \frac{4}{3}$, $z = \frac{4}{3}$

Solve: \[ x + y + z = 6, \quad 2x + y + 3z = 10, \quad x + 2y + 3z = 11 \]

Solution (Gauss-Jordan):

Augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 2 & 1 & 3 & 10 \\ 1 & 2 & 3 & 11 \end{bmatrix} \]

Step 1: Eliminate first column below pivot: \[ R_2 \to R_2 - 2R_1, \quad R_3 \to R_3 - R_1 \] \[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & -1 & 1 & -2 \\ 0 & 1 & 2 & 5 \end{bmatrix} \]

Step 2: Eliminate second column above and below pivot: \[ R_3 \to R_3 + R_2, \quad R_1 \to R_1 + R_2 \] \[ \begin{bmatrix} 1 & 0 & 2 & 4 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 3 & 3 \end{bmatrix} \]

Step 3: Normalize third row: \[ R_3 \to \frac{1}{3}R_3 \] \[ \begin{bmatrix} 1 & 0 & 2 & 4 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 1 & 1 \end{bmatrix} \]

Step 4: Eliminate third column from rows 1 and 2: \[ R_1 \to R_1 - 2R_3, \quad R_2 \to R_2 + R_3 \] \[ \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} \]

Solution: $x=2, y=1, z=1$

System: \[ x + 2y = 3, \quad 2x + 3y = 7 \]

Solution (Gauss-Jordan):

Augmented matrix: \[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 7 \end{bmatrix} \]

Step 1: Eliminate first column in second row: \[ R_2 \to R_2 - 2R_1 \] \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & 1 \end{bmatrix} \]

Step 2: Normalize second row: \[ R_2 \to -R_2 \] \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & -1 \end{bmatrix} \]

Step 3: Eliminate second column from first row: \[ R_1 \to R_1 - 2R_2 \] \[ \begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & -1 \end{bmatrix} \]

Solution: $x=5, y=-1$ (*Note: check original numbers; if intended solution is $x=1, y=1$, correct constants accordingly*)

System: \[ x + y + z = 4, \quad x - y + z = 2, \quad 2x + y - z = 3 \]

Solution (Gauss-Jordan):

Augmented matrix: \[ [A|\mathbf{b}] = \begin{bmatrix} 1 & 1 & 1 & 4 \\ 1 & -1 & 1 & 2 \\ 2 & 1 & -1 & 3 \end{bmatrix} \]

Step 1: Eliminate first column below pivot: \[ R_2 \to R_2 - R_1, \quad R_3 \to R_3 - 2R_1 \] \[ \begin{bmatrix} 1 & 1 & 1 & 4 \\ 0 & -2 & 0 & -2 \\ 0 & -1 & -3 & -5 \end{bmatrix} \]

Step 2: Normalize second row: \[ R_2 \to -\frac{1}{2}R_2 \] \[ \begin{bmatrix} 1 & 1 & 1 & 4 \\ 0 & 1 & 0 & 1 \\ 0 & -1 & -3 & -5 \end{bmatrix} \]

Step 3: Eliminate second column in other rows: \[ R_1 \to R_1 - R_2, \quad R_3 \to R_3 + R_2 \] \[ \begin{bmatrix} 1 & 0 & 1 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & -3 & -4 \end{bmatrix} \]

Step 4: Normalize third row: \[ R_3 \to -\frac{1}{3}R_3 \] \[ \begin{bmatrix} 1 & 0 & 1 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{4}{3} \end{bmatrix} \]

Step 5: Eliminate third column in first row: \[ R_1 \to R_1 - R_3 \] \[ \begin{bmatrix} 1 & 0 & 0 & 5/3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4/3 \end{bmatrix} \]

Solution: $x = 5/3, y = 1, z = 4/3$

Solve: \[ \begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}5 \\ 11\end{bmatrix} \]

Solution:

Matrix $A = \begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix}$, $\mathbf{b} = \begin{bmatrix}5 \\ 11\end{bmatrix}$.

Compute determinant: \[ \det(A) = 2\cdot4 - 1\cdot3 = 8-3 = 5 \neq 0 \]

Compute inverse: \[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}4 & -1 \\ -3 & 2\end{bmatrix} = \frac{1}{5} \begin{bmatrix}4 & -1 \\ -3 & 2\end{bmatrix} \]

Compute solution: \[ \mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{5} \begin{bmatrix}4 & -1 \\ -3 & 2\end{bmatrix} \begin{bmatrix}5 \\ 11\end{bmatrix} = \begin{bmatrix}1 \\ 3\end{bmatrix} \]

Solve: \[ \begin{bmatrix}1 & 2 \\ 3 & 1\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}5 \\ 10\end{bmatrix} \]

Solution:

Matrix $A = \begin{bmatrix}1 & 2 \\ 3 & 1\end{bmatrix}$, $\mathbf{b} = \begin{bmatrix}5 \\ 10\end{bmatrix}$.

Compute determinant: \[ \det(A) = 1\cdot1 - 2\cdot3 = 1-6 = -5 \neq 0 \]

Compute inverse: \[ A^{-1} = \frac{1}{-5} \begin{bmatrix}1 & -2 \\ -3 & 1\end{bmatrix} = \begin{bmatrix}-\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5}\end{bmatrix} \]

Compute solution: \[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix}-\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5}\end{bmatrix} \begin{bmatrix}5 \\ 10\end{bmatrix} = \begin{bmatrix}0 \\ 2.5\end{bmatrix} \]

Solve: \[ \begin{bmatrix}2 & -1 & 0 \\ 1 & 1 & 1 \\ 3 & 0 & 2\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 6 \\ 8\end{bmatrix} \]

Solution:

Let $A = \begin{bmatrix}2 & -1 & 0 \\ 1 & 1 & 1 \\ 3 & 0 & 2\end{bmatrix}$, $\mathbf{b} = \begin{bmatrix}1 \\ 6 \\ 8\end{bmatrix}$.

Compute $\det(A)$: \[ \det(A) = 2(1\cdot2 - 1\cdot0) - (-1)(1\cdot2 - 1\cdot3) + 0 = 4 - (-1)(-1) + 0 = 3 \neq 0 \]

Compute adjoint and inverse: \[ A^{-1} = \frac{1}{3} \begin{bmatrix}2 & 1 & -1 \\ -2 & 4 & -1 \\ 1 & -2 & 3\end{bmatrix} \]

Compute solution: \[ \mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{3} \begin{bmatrix}2 & 1 & -1 \\ -2 & 4 & -1 \\ 1 & -2 & 3\end{bmatrix} \begin{bmatrix}1 \\ 6 \\ 8\end{bmatrix} = \begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix} \]

System: \[ x + y + z = 3, \quad 2x + y + 3z = 7 \] Augmented matrix rank = 2, number of unknowns = 3 $\implies$ infinitely many solutions.

System: \[ x + y = 2, \quad 2x + 2y = 4 \] Rank$(A) = 1$, rank$([A|\mathbf{b}]) = 1$, unknowns = 2 $\implies$ infinitely many solutions.

Solve: \[ x + 2y = 5, \quad 3x + y = 4 \] \(\det(A) = 1\cdot1 - 3\cdot2 = -5 \neq 0\) \[ x = \frac{\det \begin{bmatrix} 5 & 2 \\ 4 & 1 \end{bmatrix}}{-5} = \frac{5-8}{-5} = \frac{-3}{-5} = \frac{3}{5}, \quad y = \frac{\det \begin{bmatrix}1 & 5 \\ 3 & 4\end{bmatrix}}{-5} = \frac{4-15}{-5} = \frac{-11}{-5} = \frac{11}{5} \]

System: \[ x + y + z = 0, \quad 2x + 3y + z = 0 \] Rank$(A) = 2$, unknowns = 3 $\implies$ dimension of solution space = 1, so infinitely many non-trivial solutions exist.

Solve: \[ x + y + z = 6, \quad 2x - y + 3z = 14, \quad x + 2y + 3z = 13 \] Compute $\det(A)$ and the determinants $\det(A_x), \det(A_y), \det(A_z)$ to find $x=2, y=1, z=3$.

System: \[ x + y + z = 3, \quad 2x + 2y + 2z = 6, \quad x - y + z = 1 \] Use row operations to reduce to REF and determine rank: - Rank$(A) = 2$, rank$([A|\mathbf{b}]) = 2$, unknowns = 3 $\implies$ infinite solutions.

System: \[ x + 2y = 0, \quad 2x + 4y = 0 \] $\det(A) = 0 \implies$ infinitely many solutions (non-trivial).