 In this video, we present the solution to question number 15 from practice exam number three from math 2270 in which case We're given a two by two matrix Here a is negative four two three and one and we're asked to compute the eigenvalues of a so the first thing When you do is compute the characteristic polynomial so phi of Lambda sub a here so the characteristic polynomial is going to equal the determinant of Lambda i minus a some people like to do the other way around a minus lambda i it doesn't make much of a difference really It'll just change the polynomial by negative one and as we're looking for the eigenvalues not the polynomial itself You can get with that with either one. I'm gonna take Lambda i minus a here for which we're treating lambda as a variable if you don't want to write lambda You could write x that's okay as well in which case if we take lambda i minus a we're taking the determinant The determinant of the matrix I'm gonna take lambda zero zero lambda and then we have to subtract from an a negative four two three and one like so For which then we're gonna get lambda plus four We're gonna get zero minus two which is a negative two we've got zero minus three Which is a negative three and then we get lambda minus one So this is what we have to compute right here. It's a two by two two determinant So we're just gonna take the the difference of the products there of the diagonals So we're gonna get lambda plus four times lambda minus one minus negative two minus three For which we're gonna have to multiply this thing out to try to factor it So foil out the lambda plus four and lambda minus one. We're gonna get a lambda squared minus lambda plus four lambda Minus four and then you get negative one times negative two times negative three that's gonna go negative six So combining like terms we get lambda squared plus three lambda Minus ten in that situation So we need factors of negative ten that fact that factors of negative ten that have to be three so we can take lambda to be plus five and Lambda lambda plus five and lambda minus two verify that just so you're sure five times negative two is a negative ten And we're gonna get that five minus two is negative three. That is the correct factorization For which then we see our eigenvalues are gonna equal lambda is negative five and positive two