The system of equations
\(x + 3y = 7\)
\(2x - y = 0\)
is solved by the row reduction method. Its augmented matrix is \(\left[\begin{array}{cc|c} 1 & 3 & 7 \\ 2 & -1 & 0 \end{array}\right]\). After the operation \(R_2 \to R_2 - 2R_1\), the second row becomes
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left[\begin{array}{cc|c} 0 & -7 & -14 \end{array}\right]\)
The row operation \(R_2 \to R_2 - 2R_1\) acts on every element of the second row, including the constant in the last column:
\(2 - 2 \times 1 = 0\)
\(-1 - 2 \times 3 = -7\)
\(0 - 2 \times 7 = -14\)
So the second row is \(\left[\begin{array}{cc|c} 0 & -7 & -14 \end{array}\right]\), that is, \(-7y = -14\), so \(y = 2\). Then \(x + 3 \times 2 = 7\) gives \(x = 1\).
Why the other options are incorrect
Option (2) \(\left[\begin{array}{cc|c} 0 & 5 & 14 \end{array}\right]\) — Incorrect. This adds \(2R_1\) in the last two places instead of subtracting it: \(-1 + 6 = 5\) and \(0 + 14 = 14\).
Option (3) \(\left[\begin{array}{cc|c} 0 & -7 & 14 \end{array}\right]\) — Incorrect. The constant must also be reduced: \(0 - 14 = -14\), not 14.
Option (4) \(\left[\begin{array}{cc|c} 0 & -5 & -14 \end{array}\right]\) — Incorrect. The middle element is \(-1 - 6 = -7\), not \(-5\).