# Full rank of a matrix

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1. Sep 21, 2015

### squenshl

1. The problem statement, all variables and given/known data
Show that the matrix $A$ is of full rank if and only if $ad-bc \neq 0$ where $$A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$$

2. Relevant equations

3. 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 $n = 2$.

2. Sep 22, 2015

### geoffrey159

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 ?

3. Sep 22, 2015

### squenshl

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

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

4. Sep 22, 2015

### geoffrey159

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

5. Sep 22, 2015

Cheers!!!