Matrix Rank
Linear Algebra
Intro: We row-reduce to echelon/RREF, count pivot positions, and identify dependent columns. Works for rectangular matrices.
Worked example
- $\mathbf{A}=\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 0 & 1 & 1 \end{bmatrix}$
- Start: $$A=\begin{pmatrix}1&2&3\\2&4&6\\0&1&1\end{pmatrix}.$$
- Eliminate below Row1, Col1 pivot: $$R_2\leftarrow R_2-2R_1\Rightarrow \begin{pmatrix}1&2&3\\0&0&0\\0&1&1\end{pmatrix}.$$
- Swap to expose the next pivot: $$R_2\leftrightarrow R_3\Rightarrow \begin{pmatrix}1&2&3\\0&1&1\\0&0&0\end{pmatrix}.$$
- Clear above the pivot in column 2: $$R_1\leftarrow R_1-2R_2\Rightarrow \begin{pmatrix}1&0&1\\0&1&1\\0&0&0\end{pmatrix}.$$
- RREF shows pivots in columns 1 and 2. Therefore $$\boxed{\operatorname{rank}(A)=2}.$$
- Column dependence note: Column 3 = Column 1 + Column 2 (since the RREF reads $x_1 + x_2 - x_3 = 0$).
- $\mathbf{A}=\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$
- Already in RREF: $$\begin{pmatrix}1&0&2\\0&1&3\end{pmatrix}.$$
- Pivots in columns 1 and 2; column 3 is free (dependent).
- Hence $$\boxed{\operatorname{rank}(A)=2}.$$
- $\mathbf{A}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
- All rows are zero; no pivots appear in RREF.
- Therefore $$\boxed{\operatorname{rank}(A)=0}.$$
FAQs
Rectangular matrices?
Yes. Rank is at most min(m,n) for an m×n matrix.
How do pivots relate to independence?
Pivot columns (in the original matrix) form a linearly independent set; non-pivot columns are linear combinations of pivot columns.
Does scaling a row change rank?
No. Elementary row operations (swap, scale by nonzero, add multiple of a row to another) preserve rank.
Rank and solutions?
For Ax=b, compare rank(A) and rank([A|b]): equal ranks → consistent; if rank(A)=rank([A|b])=n, unique solution; if equal but < n, infinitely many; if unequal, inconsistent.
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How this calculator works
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