For a system of linear equations in the augmented matrix form $$\left[\begin{array}{rrrrr|r}1 & 4 & 3 & 4 & 5 & 0 \0 & 0 & 1 & 2 & 3 & -1 \0 & 4 & -7 & 2 & 5 & 2 \0 & -4 & 1 & 1 & 2 & 5\end{array}\right]$$ what is the next step that should be done to get the system towards row echelon form, following the rules from the TOPIC 5 video?

Question

BlackTom AI · Accepted Answer

A pivot is needed in column 2 immediately below the first row, so swap R2 with a row that has a nonzero entry in column 2 (R3 or R4).

1. Relevant knowledge points and concepts
- Row echelon form requires a leading nonzero entry (pivot) in each nonzero row, with each pivot to the right of the pivot in the row above.
- Elementary row operations: swap rows, multiply a row by a nonzero scalar, add a multiple of one row to another.
- To create the second pivot, we must have a nonzero entry in column 2 of row 2 (a row below R1). If R2 has a zero in that column, we should swap R2 with a lower row that has a nonzero entry in column 2.

2. Complete reasoning step by step
- Given rows:
  R1 = [1  4   3  4  5 | 0]
  R2 = [0  0   1  2  3 |-1]
  R3 = [0  4  -7  2  5 | 2]
  R4 = [0 -4   1  1  2 | 5]
- Pivot 1 is at R1, column 1 (good). The next pivot should be in column 2 of row 2. But R2 has a 0 in column 2, so R2 cannot be the second pivot row.

3. Intermediate calculations / example swap
- Swap R2 and R3 (one valid choice):
  New R2 = R3 = [0 4 -7 2 5 | 2]
  New R3 = R2 = [0 0 1 2 3 |-1]
  Now column 2 of row 2 is 4 (nonzero), so we can proceed to scale or eliminate below it.

4. Why the correct answer is correct
- Swapping R2 with R3 or R4 moves a nonzero entry into the (2,2) position, creating the required second pivot and putting the matrix on track for row echelon form. This follows standard pivot selection rules.

5. Why the other options are incorrect (brief)
- a (R1 → R1 − R3): changes R1 unnecessarily and does not create a pivot in row 2 column 2.
- c (R3 → R3 + 7R2): R2 has 0 in column 2, so adding multiples of R2 will not change column 2 of R3; it won’t create the needed pivot in row 2.
- d (R1 → R1 + R4): alters the first pivot row for no reason and doesn’t create a (2,2) pivot.
- e (R4 → R4 − R2): since R2 has 0 in column 2, column 2 of R4 remains −4; it doesn’t place a nonzero in row 2.
- f (R4 → R4 + R3) and g (R3 → R4 + R3): these combine lower rows but do not move a nonzero into the (2,2) position in row 2.

6. Conclusion
Swapping R2 with R3 (or with R4) is the correct next step to obtain a nonzero pivot in column 2 of row 2. This confirms answer b.

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