Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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

2 Answers 2

up vote 2 down vote accepted

Perhaps one way to understand -- Look up Cramer's Rule, e.g. http://en.wikipedia.org/wiki/Cramer%27s_rule 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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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