# 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 $0$?

• No, we're in the situation where the conditions of compatibilty are not satisfied: you can't have both $ax+by=1$ and $cx+dy=0$ as the left-hand sides are colinear. – Bernard Jan 27 '15 at 23:03
• @Bernard your observation is good: a system can have infinite solutions when the determinant is $0$. In this case, one equation is a multiple of the other, so they're colinear. But $0$ can't be a multiple of $1$, so it won't work. Thanks :) – Guerlando OCs Jan 27 '15 at 23:08

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.

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!