Matrix invertible iff det(matrix)$\neq 0$? When we want to find the inverse of the matrix 
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$
we're searching for a matrix 
$$\begin{bmatrix}x & y \\ z & w\end{bmatrix}$$
such that 
$$\begin{bmatrix}x & y \\ z & w\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\implies$$
$$ax + cy = 1 \\bx + dy = 0 \\az + cw = 0\\bz + dw = 1$$
Finding the inverse is the same as finding the solution for the $x,y$ system and $w,z$ system, which by the cramer's rule only have an unique solution iff 
$$det \left(\begin{bmatrix}a & c \\ b & d\end{bmatrix}\right) \neq 0$$
which is the same as saying
$$det \left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right) \neq 0$$
So here we have the famous result that a matrix is invertible if its determinant is not $0$. But by cramer's rule we know that if the determinant is $0$, we can be in the situation where we could have multiple solutions for $x,y$ and $z,w$ which would imply that a matrix with determinant $0$ could have multiple inverses.
So,shouldn't it be false that a matrix is invertible if and only if its determinant is not $0$?
 A: No. Generally, you should not write a question implying you actually think you might have disproved an old and well known result without doing so quite explicitly-a bit more thought will always show you're mistaken. If your system has a solution, i.e. if your matrix has an inverse, then by multiplying on the right by any other matrix we see that the map $(x,y,z,w)\mapsto (ax+cy,bx+dy,az+cw,bz+dw)$ is surjective, so that the solution is automatically unique if it exists. 
A: No. Your question shows you don't understand the definition of the inverse of a matrix and how that leads to Cramer's Rule. By definition-actually, it's a consequence of the fact the set of all invertible m by n  matrices over the real or complex numbers under matrix multiplication is a group-the inverse of each matrix is unique. This is because the inverse of a matrix of a system of m equations in n unknowns does not exist unless a finite sequence of elementary row operations on the system reduces the matrix to reduced row echelon form.If the matrix is square i.e. it represents a system of n equations in n unknowns, it can be converted to the identity matrix. This sequence of operations is equivalent to multiplying by the inverse on either the left or the right, which means the matrix represents a system with a unique solution. 
As for Cramer's rule, the proof of the algorithm determines Cramer's rule is valid if and only if the system has a unique solution. Recall the definition of Cramer's rule: Consider a system of ''n'' linear equations for ''n'' unknowns, represented in matrix multiplication form as follows:
: Ax = b\,
where the ''n'' by ''n'' matrix  A  has a nonzero determinant, and the vector  x = $${x_1,x_2,.....x_n}$$ is the column vector of the variables.
Then the theorem states that in this case the system has a unique solution, whose individual values for the unknowns are given by:
xi = $$\frac{\det(A_i)}{\det(A)} \qquad i = 1, \ldots, n \ $$
where  $$A_i$$  is the matrix formed by replacing the ''i''th column of  A  by the column vector b .
If the system does not have a unique solution, then the quotient of determinants that defines Cramer's rule can have an indeterminate form i.e. it can take the form 0/0 or 
r/0 where r is the determinant of $$A_i$$. So Cramer's rule gives no solution if the matrix doesn't have a unique solution! 
