**Toolbox:**

- The rank of a matrix A is equal to $r$ if (i) A has atleast one minor of order r that does not vanish (ii) every minor of order $r+1$ or higher vanishes. $ \rho (A)=r.$
- By elementary transformations, a matrix can by reduced to echelon ( or triangular form ) so that (i) Every row of A with all its entries being 0 occurs below every row with non-zero entries. (ii) The first non-zero entry in each non-zero row is 1. (iii) The number of zeros before the first non-zero element in a row is less than the number of zeros in the next row. The rank of the matrix in echelon form is equal to the number of non-zero rows.

Method 1

$A=\begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 3 & -7 \end{bmatrix} $ is a $3$ x $4$ matrix.

$ \therefore \rho(A) \leq 3$

The two order minors of A with $C_1$ and $C_2$ will vanish since the elements in $C_1$ and $C_2$ are proportional.

Consider the 3rd order minor

$ \begin{vmatrix} 1 & -1 & 3 \\ 2 & 1 & -2 \\ 3 & 3 & -7 \end{vmatrix} = 1(-7+6)+1(-14+6)+3(6-3)$

$ = -1-8+9=0$

Consider the 3rd order minor

$ \begin{vmatrix} 2 & -1 & 3 \\ 4 & 1 & -2 \\ 6 & 3 & -7 \end{vmatrix} = 2(-7+6)+1(-28+12)+3(12-6)$

$ -2-16+18=0$

All the 3rd order minors of A are zero.

$ \therefore \rho(A) \leq 2$

Consider the 2nd order minor $ \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = 2+4=6 \neq 0$

Since at least one 2nd order minor $ \neq 0, \rho(A) = 2$

Method 2

$A=\begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 3 & -7 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 6 & -16 \end{bmatrix} R_2 \rightarrow R_2-2R_1$

$ R_3 \rightarrow R_3-3R_1$

$ \sim \begin{vmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 0 & 0 \end{vmatrix} R_3 \rightarrow R_3 - 2R_2$

The last equivalent matrix is in echelon form with 2 nonzero rows.

$ \therefore \rho(A) = 2$