Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with matrix-equations
Search options answers only
not deleted
user 3761
This question is about equations where the unknown itself is a matrix such as Sylvester or Riccati equations. For systems of linear equations (where the unknown is a vector), use "linear-system".
8
votes
Compute $x = B^{-1}(2A+I)(C^{-1}+A)b$ without calculating matrix inverses
As was mentioned in the comment, calculating $x=M^{-1}y$ is equivalent to solving $Mx=y$. Here is the full solution:
First, you can reformulate the equation to:
$Bx=(2A+I)(C^{-1}+A)b$, and by defini …
- 402