Question

Let A be a 2x2 matrix whose eigenvalues are 2 and -3. What is the determinant of A^3 + 2A^2- 5A + 3i, where i is the identity matrix.

Answer 1

this question is just too awesome isn't it?

Answer 2

actually pretty easy...try diagonalization

Answer 3

I don't see how that will work out....

Answer 4

From the characteristic poly, I can find detA = 6
I also know det(A-2i) = det(A+3i) = 0

Answer 5

you don't need that

Answer 6

write

\[A=PDP^{-1}=P\left[\begin{matrix}2 & 0\\ 0 & -3\end{matrix}\right]P^{-1}\]

Answer 7

I got that as well, though I don't know what to do with it...

Answer 8

recall that



\[A^n=PD^nP^{-1}=P\left[\begin{matrix}2 & 0\\ 0 & -3\end{matrix}\right]^nP^{-1}\]

\[=P\left[\begin{matrix}(2)^n & 0\\ 0 & (-3)^n\end{matrix}\right]P^{-1}\]

Answer 9

\[ A^3 + 2A^2- 5A + 3I\]

\[= (PDP^{-1})^3 + 2 (PDP^{-1})^2- 5 (PDP^{-1}) + 3 (PP^{-1})\]

\[= P[D^3 + 2 D^2- 5 D + 3 ]P^{-1}\]

Answer 10

forgot my I

\[= P[D^3 + 2 D^2- 5 D + 3I ]P^{-1}\]

Answer 11

so det[A^3 + 2A^2- 5A + 3i] = det(P) + det [D^3 + 2D^2 - 5D + 3i] + det(P^-1)?

Answer 12

\[Det\left(P[D^3 + 2 D^2- 5 D + 3I ]P^{-1}\right)\]

\[=Det(P)\cdot Det\left[D^3 + 2 D^2- 5 D + 3 I]\right]\cdot Det(P^{-1})\]

Answer 13

o right lol, multiply not add, but also...

Answer 14

the P and P^-1 cancel out leaving with det[D^3 + 2D^2 - 5D + 3i] which is very obvious to figure out at that point...

Answer 15

yep

Answer 16

thx a lot, been pointed at the wrong direction for a long time now lol, thank goodness you are here :D