# 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 ?

Mar 19, 2017

$\left(a , b\right) = \left(- 7 , - 11\right)$

#### Explanation:

Given:

$M = \left(\begin{matrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{matrix}\right)$

We find:

${M}^{2} = \left(\begin{matrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{matrix}\right) \left(\begin{matrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{matrix}\right) = \left(\begin{matrix}8 & 8 & 5 \\ 8 & 7 & 8 \\ 13 & 8 & 8\end{matrix}\right)$

${M}^{3} = \left(\begin{matrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{matrix}\right) \left(\begin{matrix}8 & 8 & 5 \\ 8 & 7 & 8 \\ 13 & 8 & 8\end{matrix}\right) = \left(\begin{matrix}42 & 31 & 29 \\ 45 & 39 & 31 \\ 53 & 45 & 42\end{matrix}\right)$

Then:

${M}^{3} - 3 {M}^{2} = \left(\begin{matrix}42 & 31 & 29 \\ 45 & 39 & 31 \\ 53 & 45 & 42\end{matrix}\right) - \left(\begin{matrix}24 & 24 & 15 \\ 24 & 21 & 24 \\ 39 & 24 & 24\end{matrix}\right) = \left(\begin{matrix}18 & 7 & 14 \\ 21 & 18 & 7 \\ 14 & 21 & 18\end{matrix}\right)$

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

${M}^{3} - 3 {M}^{2} - 7 M = \left(\begin{matrix}18 & 7 & 14 \\ 21 & 18 & 7 \\ 14 & 21 & 18\end{matrix}\right) - \left(\begin{matrix}7 & 7 & 14 \\ 21 & 7 & 7 \\ 14 & 21 & 7\end{matrix}\right) = \left(\begin{matrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{matrix}\right)$

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

${x}^{3} - 3 {x}^{2} - 7 x - 11 = 0$

Mar 19, 2017

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 ($m a t h b f e$'s) such that:

$A m a t h b f e = \lambda m a t h b f e \implies \left(A - \lambda I\right) m a t h b f e = m a t h b f 0$

And to avoid the null solution, ie $m a t h b f e = m a t h b f 0$, we force $\left(A - \lambda I\right)$ to be singular, ergo its determinant is zero.

So:

$\det \left(\begin{matrix}1 - \lambda & 1 & 2 \\ 3 & 1 - \lambda & 1 \\ 2 & 3 & 1 - \lambda\end{matrix}\right) = 0$

$\implies \left(1 - \lambda\right) \cdot \det \left(\begin{matrix}1 - \lambda & 1 \\ 3 & 1 - \lambda\end{matrix}\right) - 1 \cdot \det \left(\begin{matrix}3 & 1 \\ 2 & 1 - \lambda\end{matrix}\right) + 2 \cdot \det \left(\begin{matrix}3 & 1 - \lambda \\ 2 & 3\end{matrix}\right) = 0$

$\implies \left(1 - \lambda\right) \cdot \left({\left(1 - \lambda\right)}^{2} - 3\right) - 1 \cdot \left(3 \left(1 - \lambda\right) - 2\right) + 2 \cdot \left(9 - 2 \left(1 - \lambda\right)\right) = 0$

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.