If A is invertible, then so is A^3 +I True of false? why?
For a matrix, A^3 = A*A*A. Since A^3 is invertible, then A*A*A*B = I for some matrix B. Matrix multiplication is associative even though it isn't commutative, therefore A*A*A*B = A*A*(A*B) = I so define the matrix C = A*B. Now, A*A*C = A*A*A*B = A^3*B = I so A^2 is also invertible; its inverse is C.
what happens if you let \[A=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]
the A I gave above is invertible is \(A^3+I\) invertible?
oh sorry i did find the inverse but i am still kind of lost...the person above gave me an explanation i'm trying to understand it
if \(A=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\) then \(A^3+I\) \[=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]^3+\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]+\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\]
im cnfused this also makes sense. im trying to go over my notes but i cannot find anything
you are wanting to know if \(A^3+I\) is invertible given that A is invertible...correct.
yes
then the matrix I gave above shows that \(A^3+I\) might not be invertible. (the zero matrix is not invertible)
oh ok. i worked it how using ur example n it doesn't have an inverse. correct me is im wrong....for A^3 +2A=I using the same matrix u provided it would also show t be false because the end matrix on the left and right would be different right?
A^3 +2A=I is false
yes...using the same A as above
thanks i really appreciate it your help i understand.
Join our real-time social learning platform and learn together with your friends!