# 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

#### 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

#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

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

See below

#### Explanation:

A beginner's approach.

We get the characteristic equation from the basic idea that for matrix

And to avoid the null solution, ie

So:

Which **eventually** simplifies to:

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.