Q.Show that the rank of the matrix $\begin{bmatrix}1 & 2 & -1\\3 & -1 & 2\\1 & -2 & 3\\1 & -1 & 1\end{bmatrix}$ is 3.

An elementary row (column) operation on a matrix is one of three moves: (i) interchange two rows/columns ($R_i\leftrightarrow R_j$); (ii) multiply a row/column by a non-zero scalar ($R_i\to\lambda R_i$); (iii) add to a row/column a non-zero scalar multiple of another row/column ($R_i\to R_i+\lambda R_j$). Two matrices related by a sequence of such operations are called equivalent, written $A\sim B$ -- an elementary transformation changes the matrix's appearance but never the information (rank, solution set) it encodes.

Row-echelon form. A non-zero matrix $E$ is in row-echelon form if (i) every zero row sits below every non-zero row, (ii) the first non-zero entry of each row (its pivot) lies strictly to the right of the pivot in the row above, and (iii) every entry below a pivot, in its own column, is zero. Any matrix can be driven to this form by repeated pivoting: make the current pivot entry non-zero (swapping rows if needed), then use row operations to zero out everything below it, and move to the next row.

Rank. The rank $\rho(A)$ of a matrix $A$ is the order of the largest square sub-matrix of $A$ whose determinant is non-zero (equivalently: the largest $r$ for which some $r\times r$ minor is non-zero, while every minor of order $r+1$ and above vanishes). Basic facts: $\rho(A)\ge1$ once $A$ has a non-zero entry; $\rho(I_n)=n$; for an $m\times n$ matrix, $\rho(A)\le\min\{m,n\}$; and a square matrix of order $n$ is invertible exactly when $\rho(A)=n$.

Theorem (rank via echelon form). The rank of a non-zero matrix equals the number of non-zero rows in any row-echelon form of it -- this is far faster than hunting for the largest non-vanishing minor by hand, especially for a large matrix, since every entry below a pivot is already zero and so contributes nothing extra to a minor. …