In one line
Use row operations to make the matrix upper-triangular in a staircase pattern (echelon form); the rank is the number of non-zero rows.
Explained simply
Think of a staircase. Each step starts further right than the one above, and all-zero rows sit at the bottom like the floor. Once a matrix is in this shape, counting the steps (non-zero rows) gives the rank immediately, with no determinants needed.
Technically, a matrix is in row echelon form if (i) all zero rows are at the bottom, (ii) the first non-zero entry (leading entry) of each row is to the right of the leading entry of the row above, and (iii) all entries below a leading entry are zero. Its rank equals the number of non-zero rows.
How it works
- Get a non-zero entry (preferably 1) at the top-left, swapping rows if needed.
- Use \(R_i\to R_i-kR_1\) to make every entry below it zero.
- Move to the second row and second column (or the next column with a non-zero entry) and repeat.
- Continue until the staircase is complete.
- Count the non-zero rows.
Key formula
\(\rho(A)=\) number of non-zero rows in the echelon form of \(A\).
Quick example
\(A=\begin{bmatrix}1&2&3\\2&4&6\\1&1&1\end{bmatrix}\).
\(R_2\to R_2-2R_1\), \(R_3\to R_3-R_1\): \(\begin{bmatrix}1&2&3\\0&0&0\\0&-1&-2\end{bmatrix}\).
\(R_2\leftrightarrow R_3\): \(\begin{bmatrix}1&2&3\\0&-1&-2\\0&0&0\end{bmatrix}\).
Two non-zero rows, so \(\rho(A)=2\). This agrees with the minor method.
Common mistakes
- Doing several operations in one step that depend on each other (e.g. \(R_2\to R_2-R_3\) and \(R_3\to R_3-R_2\) together).
- Counting non-zero columns instead of rows.
- Arithmetic slips; recheck each new row, because one error changes the rank.
How MU asks it
Standard 5/6-mark part: "Find the rank of \(A=\begin{bmatrix}\dots\end{bmatrix}\) by reducing it to echelon form" for a \(3\times3\) matrix (the syllabus limits these to \(3\times3\)). Write every operation over the arrow, keep the matrix layout neat, and conclude "Number of non-zero rows = 2, hence \(\rho(A)=2\)".