Option C. -2R1 + R2 will reduce the x element in row 2 to 0 in the given matrix.
Explanation:
Step 1:
The values on the second row of the given matrix are 2, 1, and 0. The 2 is multiplied with x, the 1 is multiplied with y and the 0 is multiplied with z.
So the x element in the second row of the given matrix is 2.
We must find an operation such that this value becomes 0.
Step 2:
The x element in row 1 (R1); 1,
The x element in row 2 (R2); 2, and
The x element in row 3 (R3); 2.
So we substitute these values in the given operations.
Option A. 2R1 + R2 = 2(1) + 2 = 4,
Option B. 2R3 - R2 = 2(2) - 2 = 2,
Option C. -2R1 + R2 = -2(1) + 2 = 0,
Option D. R1 + R2 = 1 + 2 = 3.
Step 3:
Of the given operation, only option C attains a value of 0 while the others do not. So option C is the answer.