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Doubt in one of the options in question about row equivalence

Step-by-step mathematics solution: Doubt in one of the options in question about row equivalence

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1. What the problem is asking

We are given a (4\times4) matrix (A) that is row‑equivalent to

[ B=\begin{bmatrix} 1&0&2&0\[2pt] 0&1&1&0\[2pt] 0&0&0&1\[2pt] 0&0&0&0 \end{bmatrix}. ]

Four statements are offered; we have to decide which one must be true for every matrix (A) that is row‑equivalent to (B).

The statements are

Letter Claim
A “(B) is not the reduced row‑echelon form (RREF) of (A).”
B “(\det(A)\neq0).”
C “(\dim\bigl(\operatorname{Null}(A)\bigr)=3).”
D “If the system (A\mathbf x=\mathbf b) is consistent for some non‑zero (\mathbf b), then it has infinitely many solutions.”

We must examine each claim using the information that (A) and (B) are row‑equivalent.

2. Preliminary facts about row‑equivalence

  • Row‑equivalent matrices have the same rank.
  • The rank of a matrix equals the number of leading‑1’s (pivots) in its RREF.
  • For an (m\times n) matrix, [ \text{nullity}(A)=n-\operatorname{rank}(A)\qquad\text{(Rank–Nullity Theorem)}. ]
  • The RREF of a matrix is unique. If a matrix (B) is already in RREF, then (B) is the RREF of every matrix row‑equivalent to it.

So we first determine the rank of (B) (hence of (A)) and check whether (B) itself satisfies the definition of RREF.

3. Is (B) in reduced row‑echelon form?

Recall the definition of RREF:

  1. All zero rows (if any) are at the bottom.
  2. The first non‑zero entry in each non‑zero row (the leading entry) is a 1.
  3. Each leading 1 is the only non‑zero entry in its column.
  4. The leading 1 of a lower row lies to the right of the leading 1 of the row above it.

Check (B) row by row.

Row Leading entry Column of leading 1
1 1 1
2 1 2
3 1 4
4 (zero row) –
  • Condition 1: the zero row is indeed the last row.
  • Condition 2: each non‑zero row starts with a 1.
  • Condition 3: the columns containing the leading 1’s are columns 1, 2, 4. In each of those columns the only non‑zero entry is the leading 1 itself. (Column 3 has non‑zero entries, but it does not contain a leading 1, which is allowed.)
  • Condition 4: the leading‑1 columns progress strictly to the right: (1<2<4).

All four conditions are satisfied, so (B) **is already in reduced row‑echelon form**. Because the RREF of a matrix is unique, the RREF of any matrix row‑equivalent to (B) is precisely (B) itself.

Consequently, statement A (“(B) is not the RREF of (A)”) is false.

4. Rank, determinant, and nullity

From the RREF we see three pivots (in columns 1, 2, 4). Hence

[ \operatorname{rank}(A)=\operatorname{rank}(B)=3. ]

4.1 Determinant (Statement B)

A square matrix is invertible (and thus has non‑zero determinant) iff its rank equals its size. Here (\operatorname{rank}(A)=3<4), so (A) is singular and

[ \det(A)=0. ]

Therefore B (“(\det(A)\neq0)”) is false.

4.2 Dimension of the null space (Statement C)

The nullity (dimension of the null space) is

[ \text{nullity}(A)=n-\operatorname{rank}(A)=4-3=1. ]

Thus the null space of (A) is one‑dimensional, not three‑dimensional.
Statement C is false.

5. Consistency and number of solutions (Statement D)

Because (\operatorname{rank}(A)=3<4), the system (A\mathbf x=\mathbf b) (if consistent) will always have one free variable. A free variable produces infinitely many solutions.

To see this explicitly, solve the equivalent system (B\mathbf x=\mathbf b):

[ \begin{cases} x_1+2x_3 = b_1,\[2pt] x_2 + x_3 = b_2,\[2pt] x_4 = b_3,\[2pt] 0 = b_4 . \end{cases} ]

  • Consistency forces (b_4=0); the other three components of (\mathbf b) can be arbitrary.
  • Variable (x_3) does not appear as a leading variable, so we may choose it freely.
  • Once a value for (x_3) is selected, the equations determine (x_1, x_2, x_4) uniquely.

Hence for any non‑zero (\mathbf b) with (b_4=0) the system is consistent and admits infinitely many solutions (parameterised by the free variable (x_3)).

Therefore Statement D is true.

6. Final answer

Letter True / False Reason
A False (B) is already in reduced row‑echelon form, so it is the RREF of (A).
B False (\operatorname{rank}(A)=3<4) ⇒ (\det(A)=0).
C False Nullity (=4-\operatorname{rank}=1), not 3.
D True Rank (=3) gives one free variable; any consistent system therefore has infinitely many solutions.

The only statement that must be true is D.

7. Common mistakes for this type of problem

Mistake Why it’s wrong How to avoid it
Thinking any non‑pivot column must be zero RREF only requires that pivot columns have a single 1; non‑pivot columns may contain arbitrary numbers. Verify the four RREF conditions directly; ignore the entries in free‑variable columns.
Confusing “rank = 4” with “determinant ≠ 0” A (4\times4) matrix has non‑zero determinant iff its rank is 4. Forgetting this leads to wrong conclusions about invertibility. Compute rank from the RREF first; then decide about the determinant.
Using rank = 3 to claim the null space has dimension 3 Nullity = (n - \text{rank}); for a (4\times4) matrix this is (4-3=1). Apply the Rank–Nullity Theorem explicitly.
Assuming a consistent system can have a unique solution when the number of variables exceeds the rank Uniqueness requires rank = number of variables. With rank < variables there is at least one free variable, giving infinitely many solutions. Check the relationship rank vs (n) before deciding about uniqueness vs. infinity.
**Ignoring the zero‑row condition in RREF    

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