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. \( \left\{\begin{array}{r}\frac{1}{3} x+2 y=1 \\ -2 x-4 y=6\end{array}\right. \)
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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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