**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

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

The columns $C_1$ and $C_2$ are proportional. So the 2 3rd order minors with $C_1$ and $C_2$ will vanish.

Consider the minor $ \begin{vmatrix} 1 & 3 & 4 \\ -2 & -1 & -3 \\ -1 & 7 & 6 \end{vmatrix} = 1(-6+21)-3(-12-3)+4(-14-1)$

$=15+45-60=0$

Consider the minor $ \begin{vmatrix} -2 & 3 & 4 \\ 4 & -1 & -3 \\ 2 & 7 & 6 \end{vmatrix} = -2(-6+21)-3(24+6)+4(28+2)$

$= -30-90+120=0$

Are 4 3rd order minors vanish. $\therefore \rho(A)\leq 2$

Consider the 2nd order minor $ \begin{vmatrix} -2 & 3 \\ 4 & -1 \end{vmatrix} = 2-12=-10 \neq 10$

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

Method 2

$A = \begin{bmatrix} 1 & -2 & 3 & 4 \\ -2 & 4 & -1 & -3 \\ -1 & 2 & 7 & 6 \end{bmatrix} \sim \begin{bmatrix} 1 & -2 & 3 & 4 \\ 0 & 0 & 5 & 5 \\ 0 & 0 & 10 & 10 \end{bmatrix} R_2 \rightarrow R_2+2R_1$

$ R_3 \rightarrow R_3+R_1$

$ \sim \begin{bmatrix} 1 & -2 & 3 & 4 \\ 0 & 0 & 5 & 5 \\ 0 & 0 & 0 & 0 \end{bmatrix} R_3 \rightarrow R_3 - 2R_2$

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

$ \therefore \rho(A) = 2$