Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

Let A be a 2×2 matrix, and assume that A2 = 0. Select ALL statements below which must be true. • A. A = 0 • B. (I+A)(I−A) = I • C. There is a column X such that X not= 0 and AX = 0 • D. If AX = 0, then X = 0 • E. None of the above

OpenStudy (anonymous):

hellooooooo no one is here !!!!!

OpenStudy (anonymous):

Do you have any guesses or ideas to which might be true or not true?

OpenStudy (anonymous):

sure B and C

OpenStudy (anonymous):

C is definitely true. good job on that one. However, I dont believe B is true. Let me see if I can come up with a counter example.

OpenStudy (anonymous):

OH, nvm nvm, you are right. For some reason I was thinking (I+A)^2

OpenStudy (anonymous):

yes, B is true as well.

OpenStudy (anonymous):

\[\left[\begin{matrix}-1 &1 \\ -1 & 1\end{matrix}\right] \times \left(\begin{matrix}t \\ t\end{matrix}\right)=\left(\begin{matrix}0 \\0\end{matrix}\right)\]

OpenStudy (anonymous):

what about D

OpenStudy (anonymous):

I can come up with a counter example to D for sure. One sec.

OpenStudy (anonymous):

OpenStudy (anonymous):

Basically, since c is true, that automatically implies that D is false.

OpenStudy (anonymous):

however \[\left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right] \times \left(\begin{matrix}0 \\ 0\end{matrix}\right)=\left(\begin{matrix}0 \\0\end{matrix}\right)\] so X can be zero

OpenStudy (anonymous):

yeah, the zero vector will always be a solution to the equation:\[Ax=0\]its what is called the trivial solution. However, given that A^2 = 0, if we have to solve the matrix equation Ax = 0 , we cannot conclude that x is the zero vector. This is because there are other non-zero vectors that are also solutions to Ax = 0.

OpenStudy (anonymous):

D is saying "If Ax = 0, then x must be zero", but we know that isnt true, since C is true.

OpenStudy (anonymous):

thanks a lot :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!