Full Rank Matrix: Determinant Condition | Rank-Nullity Theorem

[squenshl]

Homework Statement

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

Homework Equations

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 $n = 2$.

 

[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 ?

 

[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?

 

[geoffrey159]

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

 

[squenshl]

Cheers!