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In Exercises \( 47-60 \), solve the linear system by the Gauss-Jordan elimination method. 47. \( \left\{\begin{array}{r}x+9 y=8 \\ 2 x+8 y=6\end{array}\right. \) 48. \( \left\{\begin{array}{r}\frac{1}{3} x+2 y=1 \\ -2 x-4 y=6\end{array}\right. \)

Ask by Gordon Rowe. in the United States
Feb 25,2025

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47. \( x = -1 \), \( y = 1 \) 48. \( x = -6 \), \( y = \frac{3}{2} \)

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Below are the step‐by‐step Gauss–Jordan elimination procedures for each system. ––––––––––––––––––––––––––––––––––––––––––– Problem 47: Solve   x + 9y = 8   2x + 8y = 6 Step 1. Write the augmented matrix:   [ 1 9 | 8 ]   [ 2 8 | 6 ] Step 2. Eliminate x from the second row. Replace row 2 with (row 2 − 2·row 1):   Row 2: [2 − 2·1, 8 − 2·9, 6 − 2·8] = [0, −10, −10]   So the matrix becomes:     [ 1 9 | 8 ]     [ 0 −10 | −10 ] Step 3. Scale row 2 by dividing by –10 to make the pivot 1:   Row 2: [0, (–10)/(–10), (–10)/(–10)] = [0, 1, 1]   New matrix:     [ 1 9 | 8 ]     [ 0 1 | 1 ] Step 4. Eliminate y from the first row. Replace row 1 with (row 1 − 9·row 2):   Row 1: [1, 9 − 9·1, 8 − 9·1] = [1, 0, -1] The matrix is now in reduced row–echelon form:   [ 1 0 | -1 ]   [ 0 1 | 1 ] Solution:   x = -1, y = 1 ––––––––––––––––––––––––––––––––––––––––––– Problem 48: Solve   (1/3)x + 2y = 1   –2x – 4y = 6 Step 1. Write the augmented matrix:   [ 1/3  2 | 1 ]   [ -2  –4 | 6 ] It is often easier to work with whole numbers, so multiply the first row by 3 (this is an allowed row-equivalent operation):   Row 1 × 3: [1, 6, 3]   Now the system becomes:     [ 1 6 | 3 ]     [ -2 –4 | 6 ] Step 2. Eliminate x from row 2. Multiply row 1 by 2 and add to row 2:   2·Row 1: [2, 12, 6]   Now add to row 2:     [ -2 + 2, –4 + 12, 6 + 6 ] = [0, 8, 12]   The updated matrix is:     [ 1 6 | 3 ]     [ 0 8 | 12 ] Step 3. Scale row 2 by dividing by 8:   Row 2: [0, 8/8, 12/8] = [0, 1, 3/2]   Matrix now:     [ 1 6 | 3 ]     [ 0 1 | 3/2 ] Step 4. Eliminate y from row 1. Replace row 1 with (row 1 − 6·row 2):   Row 1: [1, 6 – 6·1, 3 – 6·(3/2)]     = [1, 0, 3 – 9]     = [1, 0, -6] The matrix in reduced row–echelon form is:   [ 1 0 | -6 ]   [ 0 1 | 3/2 ] Solution:   x = -6, y = 3/2 ––––––––––––––––––––––––––––––––––––––––––– Final Answers: 47. x = -1, y = 1 48. x = -6, y = 3/2

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To solve these systems using the Gauss-Jordan elimination method, we first write the augmented matrix for each system and then perform row operations to achieve reduced row echelon form. For problem 47, we start with the system: \[ \begin{aligned} x + 9y &= 8 \\ 2x + 8y &= 6 \end{aligned} \] The augmented matrix is: \[ \begin{bmatrix} 1 & 9 & | & 8 \\ 2 & 8 & | & 6 \end{bmatrix} \] We can multiply the first row by -2 and add it to the second row: \[ \begin{bmatrix} 1 & 9 & | & 8 \\ 0 & -10 & | & -10 \end{bmatrix} \] Next, we can simplify the second row: \[ \begin{bmatrix} 1 & 9 & | & 8 \\ 0 & 1 & | & 1 \end{bmatrix} \] Now, we replace the first row using the second: \[ \begin{bmatrix} 1 & 0 & | & -1 \\ 0 & 1 & | & 1 \end{bmatrix} \] Thus, the solution is \( x = -1 \), \( y = 1 \). For problem 48, we begin with the system: \[ \begin{aligned} \frac{1}{3} x + 2y &= 1 \\ -2x - 4y &= 6 \end{aligned} \] The augmented matrix becomes: \[ \begin{bmatrix} \frac{1}{3} & 2 & | & 1 \\ -2 & -4 & | & 6 \end{bmatrix} \] We can start by multiplying the first row by 3 to eliminate the fraction: \[ \begin{bmatrix} 1 & 6 & | & 3 \\ -2 & -4 & | & 6 \end{bmatrix} \] Adding 2 times the first row to the second: \[ \begin{bmatrix} 1 & 6 & | & 3 \\ 0 & 8 & | & 12 \end{bmatrix} \] We can simplify the second row: \[ \begin{bmatrix} 1 & 6 & | & 3 \\ 0 & 1 & | & \frac{3}{2} \end{bmatrix} \] Replacing the first row using the second row: \[ \begin{bmatrix} 1 & 0 & | & -6 \\ 0 & 1 & | & \frac{3}{2} \end{bmatrix} \] Hence, the solution is \( x = -6 \), \( y = \frac{3}{2} \).

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