Solve the system of non-homogeneous equations by a clean matrix form: 5x₁ + 20x₃ = 50 10x₂ + 20x₃ = -30 5x₁ - 10x₂ = 80

To solve the system of equations in a clean matrix form, we start by rewriting the equations in standard form: 1. \( 5x_1 + 20x_3 = 50 \) 2. \( 10x_2 + 20x_3 = -30 \) 3. \( 5x_1 - 10x_2 = 80 \) We can express these equations in matrix form \( Ax = b \): - Coefficient matrix \( A \): \[ A = \begin{bmatrix} 5 & 0 & 20 \\ 0 & 10 & 20 \\ 5 & -10 & 0 \end{bmatrix} \] - Variable vector \( x \): \[ x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \] - Constant vector \( b \): \[ b = \begin{bmatrix} 50 \\ -30 \\ 80 \end{bmatrix} \] Now, the system can be written as: \[ \begin{bmatrix} 5 & 0 & 20 \\ 0 & 10 & 20 \\ 5 & -10 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 50 \\ -30 \\ 80 \end{bmatrix} \] To solve for \( x \), we can use row reduction or another method to find \( x \). Here, let’s augment \( A \) with \( b \) and perform row reduction: \[ \left[\begin{array}{ccc|c} 5 & 0 & 20 & 50 \\ 0 & 10 & 20 & -30 \\ 5 & -10 & 0 & 80 \end{array}\right] \] **Step 1: Simplify the first row:** Divide the first row by 5: \[ \left[\begin{array}{ccc|c} 1 & 0 & 4 & 10 \\ 0 & 10 & 20 & -30 \\ 5 & -10 & 0 & 80 \end{array}\right] \] **Step 2: Eliminate \( x_1 \) from rows 2 and 3:** We already have: - Row 1: \( R_1 = [1, 0, 4 | 10] \) Now update row 3 using row 1: \[ R_3 = R_3 - 5R_1 \Rightarrow [5 - 5*1, -10 - 5*0, 0 - 5*4 | 80 - 5*10] = [0, -10, -20 | 30] \] Now the augmented matrix is: \[ \left[\begin{array}{ccc|c} 1 & 0 & 4 & 10 \\ 0 & 10 & 20 & -30 \\ 0 & -10 & -20 & 30 \end{array}\right] \] **Step 3: Simplifying row 2:** Divide the second row by 10: \[ \left[\begin{array}{ccc|c} 1 & 0 & 4 & 10 \\ 0 & 1 & 2 & -3 \\ 0 & -10 & -20 & 30 \end{array}\right] \] **Step 4: Eliminate \( x_2 \) from row 3:** Add \( 10R_2 \) to \( R_3 \): \[ R_3 = R_3 + 10R_2 \Rightarrow [0 + 0, -10 + 10, -20 + 20 | 30 - 30] = [0, 0, 0 | 0] \] The augmented matrix now looks like: \[ \left[\begin{array}{ccc|c} 1 & 0 & 4 & 10 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] \] **Step 5: Back-substitution:** From this row-reduced form, we can extract the equations: 1. \( x_1 + 4x_3 = 10 \) 2. \( x_2 + 2x_3 = -3 \) We solve these for \( x_1 \) and \( x_2 \): - From the second equation, we have: \[ x_2 = -3 - 2x_3 \] - Substitute \( x_2 \) into the first equation: \[ x_1 = 10 - 4x_3 \] The general solution based on \( x_3 \) (a free variable) is: \[ x = \begin{bmatrix} 10 - 4t \\ -3 - 2t \\ t \end{bmatrix} \] where \( t = x_3 \). Thus, the solution for the system of equations in parametric form is: \[ x_1 = 10 - 4t, \quad x_2 = -3 - 2t, \quad x_3 = t \] for any \( t \in \mathbb{R} \). Update (2024-12-03): To solve the given system of non-homogeneous equations using matrix form, we can represent the system in the form \( Ax = b \), where \( A \) is the coefficient matrix, \( x \) is the variable vector, and \( b \) is the constant vector. The system of equations is: 1. \( 5x_1 + 20x_3 = 50 \) 2. \( 10x_2 + 20x_3 = -30 \) 3. \( 5x_1 - 10x_2 = 80 \) First, we need to write this system in matrix form. ### Coefficient matrix \( A \): From the equations, we can extract the coefficients for \( x_1, x_2, x_3 \): \[ A = \begin{bmatrix} 5 & 0 & 20 \\ 0 & 10 & 20 \\ 5 & -10 & 0 \end{bmatrix} \] ### Variable vector \( x \): The variable vector \( x \) is: \[ x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \] ### Constant vector \( b \): The constant vector \( b \) is: \[ b = \begin{bmatrix} 50 \\ -30 \\ 80 \end{bmatrix} \] ### Formulating the Matrix Equation: Now, we can express the system in matrix form: \[ \begin{bmatrix} 5 & 0 & 20 \\ 0 & 10 & 20 \\ 5 & -10 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 50 \\ -30 \\ 80 \end{bmatrix} \] ### Solving the System: To solve for \( x \), we need to find the inverse of matrix \( A \) (if it exists) and multiply it by \( b \): \[ x = A^{-1}b \] Let’s calculate the inverse of matrix \( A \). ### 1. Compute the Determinant of \( A \): The determinant of \( A \) can be computed using cofactor expansion or directly: \[ A = \begin{bmatrix} 5 & 0 & 20 \\ 0 & 10 & 20 \\ 5 & -10 & 0 \end{bmatrix} \] Calculating the determinant: \[ \text{det}(A) = 5(10 \cdot 0 - 20 \cdot -10) + 0 + 20(0 - 10 \cdot 5) \] \[ = 5(0 + 200) + 0 - 1000 \] \[ = 1000 - 1000 = 0 \] Since the determinant of \( A \) is \( 0 \), the matrix is singular. This means that the system either has no solutions or infinitely many solutions. ### 2. Write the Augmented Matrix and Row Reduce: Let us set up the augmented matrix \([A|b]\): \[ \begin{bmatrix} 5 & 0 & 20 & | & 50 \\ 0 & 10 & 20 & | & -30 \\ 5 & -10 & 0 & | & 80 \end{bmatrix} \] We will use row reduction to row echelon form to analyze the solution. Performing row operations: - \( R_3 \leftarrow R_3 - R_1 \): \[ \begin{bmatrix} 5 & 0 & 20 & | & 50 \\ 0 & 10 & 20 & | & -30 \\ 0 & -10 & -20 & | & 30 \end{bmatrix} \] - Now we can take \( R_2 \) and \( R_3 \): - Scale \( R_2 \) down: \( R_2 \leftarrow \frac{1}{10} R_2 \): \[ \begin{bmatrix} 5 & 0 & 20 & | & 50 \\ 0 & 1 & 2 & | & -3 \\ 0 & -10 & -20 & | & 30 \end{bmatrix} \] - Scale \( R_3 \leftarrow R_3 + 10R_2 \): \[ \begin{bmatrix} 5 & 0 & 20 & | & 50 \\ 0 & 1 & 2 & | & -3 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \] Now we have our system of equations in reduced form: 1. \( 5x_1 + 20x_3 = 50 \) 2. \( x_2 + 2x_3 = -3 \) 3. \( 0 = 0 \) (No new information) ### Back Substitution: From \( x_2 + 2x_3 = -3 \): \[ x_2 = -3 - 2x_3 \] From \( 5x_1 + 20x_3 = 50 \): \[ 5x_1 = 50 - 20x_3 \\ x_1 = 10 - 4x_3 \] ### Conclusion: Let \( x_3 = t \) (a free variable), the solutions can be expressed as: \[ x_1 = 10 - 4t, \quad x_2 = -3 - 2t, \quad x_3 = t \] The solution set is parametrized as: \[ x = \begin{bmatrix} 10 - 4t \\ -3 - 2t \\ t \end{bmatrix}, \quad t \in \mathbb{R} \] Thus, the system has infinitely many solutions.