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This is not homework, but extends from a proof in my book.

EDIT We're given an $m \times m$ nonsingular matrix $B$.

According to the definition of an inverse, we can calculate each element of a matrix $B^{-1}$, for a given matrix $B$, such that each element is equal to an $(m-1) \times (m-1)$ determinant divided by an $m \times m$ nonzero determinant.

Could someone please elaborate on this? I'm not sure why I don't see this, so I may need a lot of explanation.

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If that's the definition of an inverse, throw the book away. The inverse $B^{-1}$ of $B$ is a matrix such that $B^{-1}B = I = BB^{-1}$. The formula that gives the elements of the inverse as the determinant of an $(m-1)\times(m-1)$ submatrix divided by the determinant of $B$ is of theoretical interest, but as a definition, that's terrible. – Daniel Fischer Aug 8 '13 at 21:56
@DanielFischer: I've tried to clarify. The above isn't supposed to define the inverse, it's supposed to show that we can calculate each entry in the inverse as a division of an $(m-1) \times (m-1)$ determinant by an $m \times m$ determinant. – Matt Groff Aug 8 '13 at 22:05
Phew. I'm relieved. – Daniel Fischer Aug 8 '13 at 22:06
@MattGroff: See my answer, that's why we can calculate the inverse like that. – Manos Aug 8 '13 at 22:06
up vote 2 down vote accepted

Perhaps one way to understand -- Look up Cramer's Rule, e.g. and think about what the rule means when you want to find $x$ that solves $Ax = e_i$ where $e_i$ is a basis vector (all 0, except 1 at position $i$ in the vector). If you understand why Cramer's Rule is true, then you're basically there -- just note that if you want to solve the equations $Ax = e_i$ for all basis vectors (i.e. giving the identity matrix when you put the basis vectors and solutions into one matrix equation, giving $AX = I$) then you will get the determinant and sub-determinant based formula for matrix inverse $X = A^{-1}$ you described.

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Hey, thanks! It's totally obvious now. I may have to review my linear algebra, seeing that I missed this at first. – Matt Groff Aug 8 '13 at 22:25

Let $R$ be a commutative ring and $B$ an $m \times m$ matrix with entries in $R$. Then $B \cdot adj(B) = adj(B) \cdot B = det(B) \cdot I_m$, where $adj(B)$ defines the adjoint of the matrix $B$ (this is the matrix with elements being quotients of determinants as you describe). If $det(B) \neq 0$ and $det(B)$ is a unit in $R$ (which is always true if $R$is actually a field), then we can divide both sides with $det(B)$ and this by definition shows that $adj(B) / det(B) = B^{-1}$.

In summary, the interesting thing is that the formula for the inverse is valid over any commutative ring as long as the determinant of the matrix is a unit in that ring.

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