1. The problem asks to find a single elementary row operation that creates a 1 in the upper left corner of the given augmented matrix without creating fractions in the first row. 2. For matrix 17.a: $$\begin{bmatrix}-3 & -1 & 2 & 4 \\ 2 & -3 & 3 & 2 \\ 0 & 2 & -3 & 1\end{bmatrix}$$ 3. The upper left corner is currently $-3$. To get a 1 there without fractions, we can multiply the first row by $-\frac{1}{3}$, but that creates fractions. 4. Instead, we can swap the first row with the second row because the second row's first element is 2, which is not 1 but can be used to create 1 by dividing by 2, which creates fractions. 5. Alternatively, we can add multiples of rows to get 1 in the first position without fractions. But since the first row has $-3$ and the second row has 2, the greatest common divisor (gcd) of 3 and 2 is 1. 6. So, perform the operation: $R_1 \leftarrow 2R_1 + 3R_2$. Calculate the new first row: $$2 \times (-3) + 3 \times 2 = -6 + 6 = 0$$ $$2 \times (-1) + 3 \times (-3) = -2 - 9 = -11$$ $$2 \times 2 + 3 \times 3 = 4 + 9 = 13$$ $$2 \times 4 + 3 \times 2 = 8 + 6 = 14$$ This gives a 0 in the first position, not 1. 7. Try $R_1 \leftarrow -\frac{1}{3} R_1$ creates fractions, so not allowed. 8. Try swapping rows 1 and 3: $$\begin{bmatrix}0 & 2 & -3 & 1 \\ 2 & -3 & 3 & 2 \\ -3 & -1 & 2 & 4\end{bmatrix}$$ First element is 0, no good. 9. Try swapping rows 2 and 3: $$\begin{bmatrix}-3 & -1 & 2 & 4 \\ 0 & 2 & -3 & 1 \\ 2 & -3 & 3 & 2\end{bmatrix}$$ No improvement. 10. Since the gcd of 3 and 2 is 1, we can create 1 by a linear combination: Perform $R_1 \leftarrow (-1) \times R_1 + R_2$: Calculate: $$-1 \times (-3) + 2 = 3 + 2 = 5$$ Not 1. 11. Try $R_1 \leftarrow R_1 + R_2$: $$-3 + 2 = -1$$ No 1. 12. Try $R_1 \leftarrow R_1 + 2 \times R_2$: $$-3 + 2 \times 2 = -3 + 4 = 1$$ Success! No fractions created. Calculate other elements: $$-1 + 2 \times (-3) = -1 - 6 = -7$$ $$2 + 2 \times 3 = 2 + 6 = 8$$ $$4 + 2 \times 2 = 4 + 4 = 8$$ 13. So the operation is: $$R_1 \leftarrow R_1 + 2R_2$$ 14. This creates a 1 in the upper left corner without fractions in the first row. Final answer: $$\boxed{R_1 \leftarrow R_1 + 2R_2}$$