1. The problem asks to perform the row operation $\frac{1}{2} R_1$ on the first row of the matrix and keep the other rows unchanged. 2. The original matrix is: $$\begin{bmatrix} 2 & 6 & 2 & 8 \\ 3 & -1 & -2 & 4 \\ 4 & 1 & 2 & 8 \end{bmatrix}$$ 3. Applying $\frac{1}{2} R_1$ means multiplying every element in the first row by $\frac{1}{2}$: $$\frac{1}{2} \times 2 = 1$$ $$\frac{1}{2} \times 6 = 3$$ $$\frac{1}{2} \times 2 = 1$$ $$\frac{1}{2} \times 8 = 4$$ 4. The second and third rows remain the same: Second row: $3, -1, -2, 4$ Third row: $4, 1, 2, 8$ 5. So the resulting matrix is: $$\begin{bmatrix} 1 & 3 & 1 & 4 \\ 3 & -1 & -2 & 4 \\ 4 & 1 & 2 & 8 \end{bmatrix}$$ 6. Assigning the variables: $a_1 = 1$, $b_1 = 3$, $c_1 = 1$, $d_1 = 4$ $a_2 = 3$, $b_2 = -1$, $c_2 = -2$, $d_2 = 4$ $a_3 = 4$, $b_3 = 1$, $c_3 = 2$, $d_3 = 8$ Final answer: $a_1=1$, $b_1=3$, $c_1=1$, $d_1=4$ $a_2=3$, $b_2=-1$, $c_2=-2$, $d_2=4$ $a_3=4$, $b_3=1$, $c_3=2$, $d_3=8$