Write the augmented matrix of the system and use the matrix method to solve the system. \[ \begin{array}{rr} x+y+z= & 210 \\ x-2 y= & 0 \\ 2 x+3 y+5 z= & 570 \\ 2 x-y+z= & 210 \end{array} \] Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. This system has exactly one solution. The solution is \( \square \) , \( \square \) , \( ) \) ). (Type an exact answer in simplified form.) B. This system has infinitely many solutions of the form \( \square \) , ,z), where \( z \) is any real number. (Type expressions using \( z \) as the variable.) C. This system has no solution.

To solve the given system of equations using the matrix method, we first need to write the augmented matrix for the system. The system of equations is: \[ \begin{array}{rcl} x + y + z & = & 210 \\ x - 2y & = & 0 \\ 2x + 3y + 5z & = & 570 \\ 2x - y + z & = & 210 \end{array} \] ### Step 1: Write the Augmented Matrix The augmented matrix for this system can be written as follows: \[ \begin{bmatrix} 1 & 1 & 1 & | & 210 \\ 1 & -2 & 0 & | & 0 \\ 2 & 3 & 5 & | & 570 \\ 2 & -1 & 1 & | & 210 \end{bmatrix} \] ### Step 2: Row Reduction Next, we will perform row operations to reduce this matrix to row echelon form. 1. Subtract Row 1 from Row 2: \[ R_2 = R_2 - R_1 \implies R_2 = [0, -3, -1 | -210] \] 2. Subtract 2 times Row 1 from Row 3: \[ R_3 = R_3 - 2R_1 \implies R_3 = [0, 1, 3 | 150] \] 3. Subtract 2 times Row 1 from Row 4: \[ R_4 = R_4 - 2R_1 \implies R_4 = [0, -3, -1 | -210] \] Now the matrix looks like this: \[ \begin{bmatrix} 1 & 1 & 1 & | & 210 \\ 0 & -3 & -1 & | & -210 \\ 0 & 1 & 3 & | & 150 \\ 0 & -3 & -1 & | & -210 \end{bmatrix} \] 4. Now, we can simplify Row 2 and Row 4: \[ R_2 = \frac{1}{-3}R_2 \implies R_2 = [0, 1, \frac{1}{3} | 70] \] \[ R_4 = R_4 - R_2 \implies R_4 = [0, 0, 0 | 0] \] Now the matrix looks like this: \[ \begin{bmatrix} 1 & 1 & 1 & | & 210 \\ 0 & 1 & \frac{1}{3} & | & 70 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \] 5. Substitute Row 2 into Row 1: \[ R_1 = R_1 - R_2 \implies R_1 = [1, 0, \frac{2}{3} | 140] \] Now the matrix looks like this: \[ \begin{bmatrix} 1 & 0 & \frac{2}{3} & | & 140 \\ 0 & 1 & \frac{1}{3} & | & 70 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \] ### Step 3: Back Substitution From the reduced matrix, we can express the variables in terms of \( z \): 1. From Row 1: \[ x + \frac{2}{3}z = 140 \implies x = 140 - \frac{2}{3}z \] 2. From Row 2: \[ y + \frac{1}{3}z = 70 \implies y = 70 - \frac{1}{3}z \] ### Conclusion The system has infinitely many solutions of the form: \[ (x, y, z) = \left(140 - \frac{2}{3}z, 70 - \frac{1}{3}z, z\right) \] Thus, the correct choice is: **B. This system has infinitely many solutions of the form \( (140 - \frac{2}{3}z, 70 - \frac{1}{3}z, z) \), where \( z \) is any real number.**