System of Linear Equations PDF

## System of Linear Equations **نظام من المعادلات الخطية** * 2x + 3y - z = 7 * -x + y - 2z = 3 * 2x - z = 9 **Ax = B** $A = \begin{bmatrix} 2 & 3 & -1 \\ -1 & 1 & -2 \\ 2 & 0 & -1 \end{bmatrix}$, $B = \begin{bmatrix} 7 \\ 3 \\ 9 \end{bmatrix}$, $x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$...

## System of Linear Equations **نظام من المعادلات الخطية** * 2x + 3y - z = 7 * -x + y - 2z = 3 * 2x - z = 9 **Ax = B** $A = \begin{bmatrix} 2 & 3 & -1 \\ -1 & 1 & -2 \\ 2 & 0 & -1 \end{bmatrix}$, $B = \begin{bmatrix} 7 \\ 3 \\ 9 \end{bmatrix}$, $x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ **كل نظام المعادلات** * Augmented Matrix [A|B] **Row Echelon Form** * Gaussian Elimination Method * طريقة جاوس للحذف **Reduced Row Echelon Form** * Gauss-Jordan * طريقة جاوس جودون **Homogeneous** **متجانس** * Ax = 0 * 2x + 3y - z = 0 * -x + y - 2z = 0 * 2x - z = 0 **Inhomogeneous** **غير متجانس** * Ax = B * 2x + 3y - z = 7 * -x + y - 2z = 3 * 2x - z = 9 --- **In Homogeneous** **غير المتجانس** - Unique Solution - حل وحيد - (One Solution) - Consistent - متوافق - Infinitely Many Solutions - عدد لانهائي من الحلول - Solutions - Consistent - متوافق - No Solution - لا يوجد حل - Inconsistent --- **Homogeneous** **المتجانس** - Trivial Solution - الحل التافه (الصغري) - Consistent - متوافق - Infinitely Many Solutions - عدد لانهائي من الحلول - Consistent - متوافق --- **Rank of a matrix** **رتبه بصعوفة** * It's the number of non-zero rows after putting the matrix in row echelon form. * هي عدد الصفوف غير جعفريه بعد وضع المصفوفة في الصورة المختزلة **Dif:** * R(A) = R(A|B) = number of variables => One Solution * عدد المجاهيل => حل واحد * R(A) < R(A|B) => No Solution * لا يوجد حل * R(A) = R(A|B) => Infinitely Many Solutions * يوجد عد ذل اشهائي من حلول --- **Ex: Solve the system of Linear Equation By Using Gauss and Gauss-Jordan Elimination Method** 2x + 4y = 3 x - 3y + 5z = 1 3x - y - z = 4 [A|B] = $\begin{bmatrix} 2 & 4 & & 3 \\ 1 & -3 & 5 & 1 \\ 3 & -1 & -1 & 4 \end{bmatrix}$ $R_1 \leftrightarrow R_2$ $\begin{bmatrix} 1 & -3 & 5 & 1 \\ 2 & 4 & & 3 \\ 3 & -1 & -1 & 4 \end{bmatrix}$ -3R₁ + R₃ $\begin{bmatrix} 1 & -3 & 5 & 1 \\ 2 & 4 & & 3 \\ 0 & 8 & -16 & 1 \end{bmatrix}$ -2R₁ + R₂ $\begin{bmatrix} 1 & -3 & 5 & 1 \\ 0 & 10 & -10 & 1 \\ 0 & 8 & -16 & 1 \end{bmatrix}$ -8R₂ + R₃ $\begin{bmatrix} 1 & -3 & 5 & 1 \\ 0 & 10 & -10 & 1 \\ 0 & 0 & -8 & -7 \end{bmatrix}$ -3/2R₃ + R₂ $\begin{bmatrix} 1 & -3 & 5 & 1 \\ 0 & 10 & 0 & \frac{11}{2} \\ 0 & 0 & -8 & -7 \end{bmatrix}$ -5/8R₃ + R₁ $\begin{bmatrix} 1 & -3 & 0 & \frac{1}{8} \\ 0 & 10 & 0 & \frac{11}{2} \\ 0 & 0 & -8 & -7 \end{bmatrix}$ 1/10R₂ $\begin{bmatrix} 1 & -3 & 0 & \frac{1}{8} \\ 0 & 1 & 0 & \frac{11}{20} \\ 0 & 0 & -8 & -7 \end{bmatrix}$ 3R₂ + R₁ $\begin{bmatrix} 1 & 0 & 0 & \frac{11}{16} \\ 0 & 1 & 0 & \frac{11}{20} \\ 0 & 0 & -8 & -7 \end{bmatrix}$ -1/8R₃ $\begin{bmatrix} 1 & 0 & 0 & \frac{11}{16} \\ 0 & 1 & 0 & \frac{11}{20} \\ 0 & 0 & 1 & \frac{7}{8} \end{bmatrix}$ By Using Gauss * z = 7/8 * y + 2z = 11/20 => y = 11/20 -2(7/8) = 5/8 * x - 3y + 5z = 1 => x = 1 + 3(5/8) - 5(7/8) = 11/16 (x, y, z) = (11/16, 5/8, 7/8) --- **Using Gauss-Jordan** $\begin{bmatrix} -3 & 5 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ * -2R₃ + R₂ $\begin{bmatrix} -3 & 5 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ * -5R₃ + R₁ $\begin{bmatrix} -3 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ * 3R₃ + R₁ $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ It can be checked by substituting the values of x, y, and z in one of the equations. --- **Solve** x + 2y + 3z = 1 -3x - 2y - z = 2 4x + 4y + 4z = 3 [A|B] = $\begin{bmatrix} 1 & 2 & 3 & 1 \\ -3 & - 2 & -1 & 2 \\ 4 & 4 & 4 & 3 \end{bmatrix}$ * 3R₁ + R₂ $\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 8 & 5 \\ 4 & 4 & 4 & 3 \end{bmatrix}$ * -4R₁ + R₃ $\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 8 & 5 \\ 0 & -4 & -8 & -1 \\ \end{bmatrix}$ * R₂ + R₃ $\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 8 & 5 \\ 0 & 0 & 0 & 4 \\ \end{bmatrix}$ 0 = 4 , Impossible The system is Inconsistent (has no solution). --- **Solve** * x - 2x₂ + x₃ + x₄ = 0 * -x₁ + 2x₂ + x₄ = 0 * 2x₁ - 4x₂ + x₃ + x₄ = 0 [A|B] = $\begin{bmatrix} 1 & - 2 & 1 & 1 & 0 \\ -1 & 2 & & 1 & 0 \\ 2 & -4 & 1 & 1 & 0 \end{bmatrix}$ * R₁ + R₂ $\begin{bmatrix} 1 & - 2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 2 & -4 & 1 & 1 & 0 \end{bmatrix}$ * -2R₁ + R₃ $\begin{bmatrix} 1 & - 2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & -1 & -1 & 0 \end{bmatrix}$ * R₂ + R₃ $\begin{bmatrix} 1 & - 2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{bmatrix}$ x₁ - 2x₂ + x₃ + x₄ = 0 Basic x₃ + 2x₄ = 0 Free x₄ = 0 * x₃ + 2x₄ = 0 => x₃ = - 2x₄ * x₁ - 2x₂ - x₄ = 0 => x₁ = 2x₂ + x₄ * x₄ = 0 The set of all solutions are (x₁, x₂, x₃, x₄) = (2s + t, s, -2t+t) let x₂ = s, x₄ = t, s, t ∈ ℝ (x₁, x₂, x₃, x₄) = (2s + t, s, -2t + t) Let s = 0, t = 1, (x₁, x₂, x₃, x₄) = (1, 0, -2, 1) --- ## Find the Values of λ where in makes the system of Equations has * x + y - z = 2 * x - y + λz = 2 * -x - y + λ²z = 4 1. **Unique Solution and Find its Solution** 2. **Has Infinitely Many Solutions** 3. **Inconsistent (No Solution)** [A|B] = $\begin{bmatrix} 1 & 1 & -1 & 2 \\ 1 & -1 & \lambda & 2 \\ -1 & -1 & \lambda^2 & 4 \end{bmatrix}$ * R₁ + R₂ $\begin{bmatrix} 1 & 1 & -1 & 2 \\ 0 & -2 & \lambda + 1 & 0 \\ -1 & -1 & \lambda^2 & 4 \end{bmatrix}$ * R₁+ R₃ $\begin{bmatrix} 1 & 1 & -1 & 2 \\ 0 & -2 & \lambda + 1 & 0 \\ 0 & 0 & \lambda^2 - 1 & 6 \end{bmatrix}$ * (-1/2)R₂ $\begin{bmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & -\frac{\lambda + 1}{2} & 0 \\ 0 & 0 & \lambda^2 - 1 & 6 \end{bmatrix}$ * -R₂ + R₁ $\begin{bmatrix} 1 & 0 & -\frac{\lambda - 1}{2} & 2 \\ 0 & 1 & -\frac{\lambda + 1}{2} & 0 \\ 0 & 0 & \lambda^2 - 1 & 6 \end{bmatrix}$ 1. **If λ ≠ 1 ≠ 0 => λ + 1 ≠ 0 => λ ∈ R - {1, - 1}** The system has a Unique Solution. (One Solution) Put λ = 2 $\begin{bmatrix} 1 & 0 & -\frac{1}{2} & 2 \\ 0 & 1 & -\frac{3}{2} & 0 \\ 0 & 0 & 3 & 6 \end{bmatrix}$ * R₃ /3 $\begin{bmatrix} 1 & 0 & -\frac{1}{2} & 2 \\ 0 & 1 & -\frac{3}{2} & 0 \\ 0 & 0 & 1 & 2 \end{bmatrix}$ * (1/2)R₃+R₁ $\begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & -\frac{3}{2} & 0 \\ 0 & 0 & 1 & 2 \end{bmatrix}$ * (3/2)R₃ + R₂ $\begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end{bmatrix}$ => z = 2 => y - 3/2 * z = 0 => y - 3/2 * 2 = 0 => y = 3 => x + y - z = 2 => x + 3 - 2 = 2 => x = 1 (x, y, z) = (1, 3, 2) 2. **Impossible (Infinitely many solutions)** 3. **The system has Inconsistent (No solution) if λ = 1**