Invertible Matrices and Linear independence If a matrix is invertible, what does this tell us application wise?
I am familiar with what it implies in regards to the properties of the matrix, i.e:
the determinant is non-zero, and for a matrix $A$, $Ax=0$ implies $x=0$.
However, during discussions in my lectures for an optimization class, a class mate always brings up non-invertible and invertible matrices and what they imply. I can't be too specific as I tend to get lost in discussion whenever this happens, but could anyone clue me in on what the significance of invertibility and linear independence (of which the definitions I am aware of) is?
Kind regards.
 A: I had a lot of trouble getting my head around this myself and apologise if anything in my answer is incorrect.
Consider the matrix expression
$$\begin{bmatrix}
3&6\\
1&2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}$$
 I can take a factor of 3 out of the 1st row
$$
3\begin{bmatrix}
1&2\\
1&2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}$$
The matrix has a zero determinant by inspection or $2-2=0$
A matrix with zero determinant is singular and has no inverse.
Notice that the 1st row is obviously a linear combination of the second row and so they are linearly dependent. This was just an example to get a feeling for what is happening. It is more appropriate to think of the determinant as being designed to test for linear dependence. Herbert Gross has excellent lectures from MIT. 
A: Hint : The Caley-hamilton-theorem states that a matrix with characteristic equation
        $$a_n\lambda^n+...+a_1\lambda+a_0=0$$
        satisfies the equation 
        $$a_nA^n+...+a_1A+a_0I=0$$
Also note, that A can be written as AI.
Now, do you see how $A^{-1}$ can be expressed by powers of $A$ , including I ?
