4
$\begingroup$

Let $A$ be an $n \times n$ invertible matrix with \begin{align} \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{array}\right)^{-1}= \left(\begin{array}{ccc} b_{11} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nn} \end{array}\right) \end{align} Is there an explicit formula for $b_{ij}$ in terms of the elements of $A$ and the determinant of $A$?

Edit: Here is a link to the different possible methods to invert a matrix http://en.m.wikipedia.org/wiki/Invertible_matrix#Methods_of_matrix_inversion.

Originally I had in mind using the blockwise inversion method but I think all of the methods require using some submatrix of $A$ and their determinants and / or inverses. I don't think it's possible to write something explicit in terms of $\{a_{ij}\}_{i=1,j=1}^{n,n}$

$\endgroup$
1
3
$\begingroup$

Hint: $A^{-1}=\frac{1}{|A|}\cdot \text{Adj}(A)$

$\text{Adj}(A)$ is the adjunct matrix of A.

$\endgroup$
3
  • $\begingroup$ Also called the adjugate matrix or the classical adjoint. $\endgroup$ – J. David Taylor Oct 27 '14 at 1:22
  • $\begingroup$ Thanks. It still seems like this would not be completely explicit but rather in terms of determinants of submatrices? e.g. $b_{11}=\frac{1}{|A|} \left|\begin{array}{ccc}a_{22} &\cdots & a_{2n} \\ \vdots & \ddots & \vdots \\ a_{n2} & \cdots & b_{nn} \end{array}\right|$. $\endgroup$ – user103828 Oct 27 '14 at 12:27
  • $\begingroup$ At the moment I don´t see a possibility to write it without the determinants of the submatrices: $ A^{-1}=\frac{1}{|A|}\cdot\begin{pmatrix} |A_{11}| & -|A_{12}| & \ldots & (-1) ^{1+n}|A_{1n}|\\ \vdots & \vdots & \vdots & \vdots \\ (-1) ^{n+1}|A_{n1}| & (-1) ^{n+2}|A_{22}| & \ldots & (-1) ^{n+n}|A_{nn}| \end{pmatrix}^T$ $\endgroup$ – callculus Oct 27 '14 at 13:19

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.