A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Let A be the matrix \[A=\left[\begin{matrix}1 & 2\\ 2 &4\end{matrix}\right]\] decide all the 2x2 matrices B such that AB=BA=0, where 0 is the zeromatrix
 2 years ago
Let A be the matrix \[A=\left[\begin{matrix}1 & 2\\ 2 &4\end{matrix}\right]\] decide all the 2x2 matrices B such that AB=BA=0, where 0 is the zeromatrix

This Question is Closed

frx
 2 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right]\left[\begin{matrix}a &b \\ b & c\end{matrix}\right]=\left[\begin{matrix}0 &0 \\ 0& 0\end{matrix}\right]\] \[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}1 &2\\ 2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0\\ 0 &0\end{matrix}\right]\] So all I can figure is that for AB=BA=0 B can be\[B=A ^{1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\]

frx
 2 years ago
Best ResponseYou've already chosen the best response.0But is there any other solution than the inverse?

frx
 2 years ago
Best ResponseYou've already chosen the best response.0I guess it could be \[\left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right]\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\] \[\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\] but so the two solutions for AB=BA=0 should be B=A^1 and B=0, am I right?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1\(A\) is not invertable...so how can \(B=A^{1}\)

frx
 2 years ago
Best ResponseYou've already chosen the best response.0It's not I guess, A is not invertable if AX=0 is the definition I remember right now, is that right?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1\(Det(A)=0\) so \(A^{1}\) does not exist

frx
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, but is there any other solution than B equals the zerovector?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1yes..there are infinitly many solutions

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1you have 2 of the solutions

frx
 2 years ago
Best ResponseYou've already chosen the best response.0So how do I show that it has infinity many solutions?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1show that \(B=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\) is a solution

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1then \(B\cdot t\) is a solution for all real numbers \(t\)

frx
 2 years ago
Best ResponseYou've already chosen the best response.0So B can be the same as what the inverse of A would have been if it existed?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1it is not really the same

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1remember you have to divide by the det(A) for the shortcut way to find the inverse

frx
 2 years ago
Best ResponseYou've already chosen the best response.0Oh I get it, it can't be the inverse since the definition says that A^1*A=I isn't that right?

frx
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you for your help! :)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.