Given matrix #M = ((1, 1, 2), (3, 1, 1), (2, 3, 1))# with characteristic equation #x^3-3x^2+ax+b = 0#, what are the values of #a# and #b# ?

2 Answers
Mar 19, 2017

Answer:

#(a, b) = (-7, -11)#

Explanation:

Given:

#M = ((1, 1, 2), (3, 1, 1), (2, 3, 1))#

We find:

#M^2 = ((1, 1, 2), (3, 1, 1), (2, 3, 1))((1, 1, 2), (3, 1, 1), (2, 3, 1)) = ((8, 8, 5), (8, 7, 8), (13, 8, 8))#

#M^3 = ((1, 1, 2), (3, 1, 1), (2, 3, 1))((8, 8, 5), (8, 7, 8), (13, 8, 8)) = ((42, 31, 29), (45, 39, 31), (53, 45, 42))#

Then:

#M^3-3M^2 = ((42, 31, 29), (45, 39, 31), (53, 45, 42))-((24, 24, 15), (24, 21, 24), (39, 24, 24)) = ((18, 7, 14), (21, 18, 7), (14, 21, 18))#

Looking at the off diagonal elements, it is clear that we want to set #a=-7# to find:

#M^3-3M^2-7M = ((18, 7, 14), (21, 18, 7), (14, 21, 18))-((7, 7, 14), (21, 7, 7), (14, 21, 7)) = ((11, 0, 0), (0, 11, 0), (0, 0, 11))#

So put #b=-11# to find that the characteristic equation is:

#x^3-3x^2-7x-11 = 0#

Mar 19, 2017

Answer:

See below

Explanation:

A beginner's approach.

We get the characteristic equation from the basic idea that for matrix #A# there exist eigenvalues (#lambda#'s) and eigenvectors (#mathbf e#'s) such that:

#A mathbf e = lambda mathbf e implies (A - lambda I) mathbf e = mathbf 0#

And to avoid the null solution, ie #mathbf e = mathbf 0#, we force #(A - lambda I) # to be singular, ergo its determinant is zero.

So:

#det ((1 - lambda, 1, 2), (3, 1 - lambda, 1), (2, 3, 1 - lambda) ) = 0#

# implies (1-lambda) * det ((1-lambda, 1),(3, 1 - lambda)) - 1 * det ((3,1),(2, 1 - lambda)) + 2 * det ((3, 1 - lambda),(2,3)) = 0#

# implies (1-lambda) * ((1-lambda)^2 - 3) - 1 * (3(1 - lambda) - 2) + 2 * (9 - 2(1-lambda)) = 0#

Which eventually simplifies to:

# lambda^3 - 3 lambda^2 - 7 lambda - 11 = 0#

Clearly this approach does use the fact you already have the cubic and quadratic terms in the bag, and involves a load of algebra that I declined to typeset, and so is weaker on those grounds alone.