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 $0$?

  • 3
    $\begingroup$ 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. $\endgroup$ – Bernard Jan 27 '15 at 23:03
  • $\begingroup$ @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 :) $\endgroup$ – 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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.