A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

hewsmike
 one year ago
Best ResponseYou've already chosen the best response.1\[A = \left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]\]\[B = \left[\begin{matrix}1 & 2 \\ 3 & 6\end{matrix}\right]\]\[C = \left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right]\]so \[AC = \left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]\left[\begin{matrix}1 &0 \\ 0 & 0\end{matrix}\right]=\left[\begin{matrix}1 &0 \\3 & 0\end{matrix}\right]\]and\[BC = \left[\begin{matrix}1 & 2 \\ 3 & 6\end{matrix}\right]\left[\begin{matrix}1 &0 \\ 0 & 0\end{matrix}\right]=\left[\begin{matrix}1 &0 \\3 & 0\end{matrix}\right]\]there are many choices here. The key issue is that C must be singular ( noninvertible ) because if it was then \[A C = B C\]becomes\[A C C^{1}= B C C^{1}\]\[A I = B I\]\[A = B\]which the question does not allow. The values in the second columns of A and/or B are irrelevant, as C is selecting only their first columns and hence the products are equal.

Omar91X
 one year ago
Best ResponseYou've already chosen the best response.0Thank you sir, I appreciate the explanation.
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.