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 2 years ago
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.
 2 years ago
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.

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Jnlucero
 2 years ago
Best ResponseYou've already chosen the best response.0the characteristic equation is x^2 +x  6 = 0 so deltaA = 6. thus A^3 + 2A^2  5A + 3i = 216 +72 +30 +3 =101

findme
 2 years ago
Best ResponseYou've already chosen the best response.0Shouldn't the characteristic equation be x^2 + x + 6 =0 so detA = 6? anyhow, I don't understand how you can just sub detA into the equation to the the determinant of A^3 + 2A^2  5A + 3i, please explain.

findme
 2 years ago
Best ResponseYou've already chosen the best response.0I was thinking of Caylay Hamilton Theroem, but I'm not quite familiar with it...

Jnlucero
 2 years ago
Best ResponseYou've already chosen the best response.0eigenvalues 2 and 3 means the roots of the characteristic equation\[\delta(AxI)\] are 2 and 3. this gives x^2 + x 6 = 0 since r +s = 1 and rs = 6. thus \[\delta(A)\] is the constant term 6. thus substitution gives the answer.

findme
 2 years ago
Best ResponseYou've already chosen the best response.0so you are saying det(A^3 + 2A^2  5A + 3i) = (detA)^3 + 2(detA)^2  5det(A) +3?

Jnlucero
 2 years ago
Best ResponseYou've already chosen the best response.0i thought so, since the problem is to find the determinant of the expression.

findme
 2 years ago
Best ResponseYou've already chosen the best response.0Is there some kind of theorem to justify what you did? Such as product rule: det(AB) = detA + detB

findme
 2 years ago
Best ResponseYou've already chosen the best response.0I simply don't see the justification behind subbing....

Jnlucero
 2 years ago
Best ResponseYou've already chosen the best response.0let me research for it....

findme
 2 years ago
Best ResponseYou've already chosen the best response.0oh and you are right about detA = 6, silly me lol
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