Change the following linear system to reduced row echelon form \[ \begin{array}{l}2 x_{1}+4 x_{2}+6 x_{3}=8 \\ x_{1}+2 x_{2}+4 x_{3}=8 \\ 3 x_{1}+6 x_{2}+9 x_{3}=12\end{array} \]

To convert the given linear system into reduced row echelon form, we start by writing the augmented matrix: \[ \begin{bmatrix} 2 & 4 & 6 & | & 8 \\ 1 & 2 & 4 & | & 8 \\ 3 & 6 & 9 & | & 12 \end{bmatrix} \] Now, we perform row operations to simplify this matrix step by step. 1. **Row Reduction**: We can start by making the first column entries below the leading 1 (in Row 2) equal to zero. To do this, we can perform the following row operations: - \( R_1 \leftarrow R_1 - 2R_2 \) - \( R_3 \leftarrow R_3 - 3R_2 \) After these operations, the matrix looks like this: \[ \begin{bmatrix} 0 & 0 & -2 & | & -8 \\ 1 & 2 & 4 & | & 8 \\ 0 & 0 & -2 & | & -12 \end{bmatrix} \] 2. **Scaling Rows**: To make the leading coefficient in the first row (which currently has all zeros, thus invalid as a leading row), let's interchange Row 1 and Row 2: \[ \begin{bmatrix} 1 & 2 & 4 & | & 8 \\ 0 & 0 & -2 & | & -8 \\ 0 & 0 & -2 & | & -12 \end{bmatrix} \] Now we simplify the second and third rows. We can divide the second row by -2: \[ R_2 \leftarrow -\frac{1}{2} R_2 \] This gives us: \[ \begin{bmatrix} 1 & 2 & 4 & | & 8 \\ 0 & 0 & 1 & | & 4 \\ 0 & 0 & -2 & | & -12 \end{bmatrix} \] Next, we will scale \( R_3 \): \[ R_3 \leftarrow -\frac{1}{2} R_3 \] Leading to: \[ \begin{bmatrix} 1 & 2 & 4 & | & 8 \\ 0 & 0 & 1 & | & 4 \\ 0 & 0 & 1 & | & 6 \end{bmatrix} \] 3. **Zeroing Out Below the Pivot**: We want to zero out entries in Row 1 and Row 3, column 3 (where we now have a leading 1 in Row 2): \[ R_1 \leftarrow R_1 - 4R_2 \] \[ R_3 \leftarrow R_3 - R_2 \] After applying these, we obtain: \[ \begin{bmatrix} 1 & 2 & 0 & | & -8 \\ 0 & 0 & 1 & | & 4 \\ 0 & 0 & 0 & | & -2 \end{bmatrix} \] Next, we have a contradiction in the third row, which leads us to understand that the system does not have a solution. In conclusion, the reduced row echelon form of the matrix would ultimately clarify that the original system is inconsistent, containing equations that describe parallel planes that do not intersect. Thus, there are no values for \(x_1\), \(x_2\), and \(x_3\) that would satisfy all three equations simultaneously.