The system of linear equations $$\begin{eqnarray}v + 2w + 3x + y + z &=& 1 \-2v -4w +6x + 2y + 2z &=& 2 \3v + 6w -3x - y - z &=& 1\end{eqnarray}$$ has row echelon form $$\left[\begin{array}{rrrrr|r}1 & 2 & 0 & 0 & 0 & -1 \0 & 0 & 3 & 1 & 1 & 2 \0 & 0 & 0 & 0 & 0 & 4\end{array}\right].$$ How many solutions does this system have?

Question

BlackTom AI · Accepted Answer

An augmented row of the form [0 0 0 0 0 | 4] is a direct contradiction (0 = 4), so the system is inconsistent and has no solutions.

1) Relevant concepts:
- Row echelon form lets us read equations directly from rows.
- The Rouché–Capelli theorem: a linear system is consistent iff the rank of the coefficient matrix equals the rank of the augmented matrix. If ranks differ the system is inconsistent (no solutions).
- Pivots (nonzero rows in coefficient part) determine rank; free variables arise when there are fewer pivots than unknowns, which would give infinitely many solutions only if the system is consistent.

2) Step-by-step reasoning and intermediate steps:
From the given row echelon augmented matrix:
Row1: 1·v + 2·w + 0·x + 0·y + 0·z = −1  ⇒ v + 2w = −1.
Row2: 0·v + 0·w + 3·x + 1·y + 1·z = 2  ⇒ 3x + y + z = 2.
Row3: 0·v + 0·w + 0·x + 0·y + 0·z = 4  ⇒ 0 = 4.

Step: The third row produces the equation 0 = 4, which is impossible. That single contradiction makes the entire system inconsistent.

Compute ranks:
- Rank of coefficient matrix A: only the first two rows of coefficients are nonzero, so rank(A) = 2.
- Rank of augmented matrix [A|b]: because the augmented column produces a nonzero third row, rank(A|b) = 3.
Since rank(A) (2) ≠ rank(A|b) (3), the system is inconsistent by Rouché–Capelli.

3) Why the correct answer is correct:
The contradiction 0 = 4 shows there is no assignment of v,w,x,y,z satisfying all three equations simultaneously. Thus there are no solutions.

4) Why the other multiple-choice options are incorrect:
- infinitely many: would require consistency with fewer pivots than variables (free variables); here the system is inconsistent, so impossible.
- three/one/two/five: these assume some consistent solution set with that many distinct solutions; linear systems either have no solution, exactly one solution, or infinitely many; discrete finite numbers >1 (except 1) are impossible — and inconsistency rules out the unique-solution case as well.
Therefore none of those apply.

Conclusion: The system has no solutions. Final answer: f. none.

类似问题

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