Which of the following matrices can NOT be obtained from the matrix $$\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$$ by a single elementary row operation?

We start with the matrix $$A = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$$ and check which option cannot be obtained by a single elementary row operation.

Option A: $$\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix}$$. Apply $$R_1 \to R_1 + R_2$$: $$(-1+1,\; 2+(-1)) = (0, 1)$$, and $$R_2$$ stays $$(1, -1)$$. This gives $$\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix}$$. Obtainable.

Option B: $$\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$$. Apply $$R_1 \leftrightarrow R_2$$ (row swap): rows get interchanged, giving $$\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$$. Obtainable.

Option C: $$\begin{pmatrix} -1 & 2 \\ -2 & 7 \end{pmatrix}$$. Row 1 is unchanged, so the operation must be on $$R_2$$. We need $$R_2 \to R_2 + kR_1$$: $$(1 + k(-1),\; -1 + 2k) = (-2, 7)$$. From the first entry: $$1 - k = -2$$, so $$k = 3$$. From the second entry: $$-1 + 6 = 5 \neq 7$$. So no value of $$k$$ works. We could also try $$R_2 \to cR_2$$: $$(c, -c) = (-2, 7)$$ gives $$c = -2$$ and $$c = -7$$, contradiction. No single elementary row operation produces this matrix. NOT obtainable.

Option D: $$\begin{pmatrix} -1 & 2 \\ -1 & 3 \end{pmatrix}$$. Row 1 unchanged, so $$R_2 \to R_2 + kR_1$$: $$(1-k,\; -1+2k) = (-1, 3)$$. From the first: $$k = 2$$. From the second: $$-1+4 = 3$$. Both consistent. Obtainable with $$R_2 \to R_2 + 2R_1$$.

Hence, the correct answer is Option C: $$\begin{pmatrix} -1 & 2 \\ -2 & 7 \end{pmatrix}$$.