# Full rank of a matrix

## Homework Statement

Show that the matrix ##A## is of full rank if and only if ##ad-bc \neq 0## where $$A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$$

## The Attempt at a Solution

Suppose that the matrix ##A## is of full rank. That is, rank ##2##. Then by the rank-nullity theorem, the
dimension of the kernel is ##0##. This implies that there exists an inverse ##A^{-1}## but this will only occur if ##ad-bc \neq 0## otherwise our matrix ##A## will be singular. On the other hand, suppose ##ad-bc \neq 0##. Hence, ##A## is nonsingular and there exists an inverse ##A^{-1}## but this will occur only when the dimension of the kernel is ##0##, that is, of rank ##n = 2##.

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Your matrix has rank 2 iff the column vectors are linearly independent.
Can you show that the determinant of the matrix is zero iff the column vectors are linearly dependent ?

Sorry $$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ if that even matters!!!!

How could I do this without using anything on determinants???

Show that ad - bc = 0 iff ##\vec u = (a,c) ## and ##\vec v = (b,d)## are linearly dependent

Cheers!!!