Hello

My daughter is doing a college assignment and I managed to help her with the majority of the question but cannot figure out the last part.

I have the characteristic equation and have made that a cubic but unsure on the M^-1 part

Thanks

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- Apr 11th 2012, 12:19 AMJane1947Help with question on matrices
Hello

My daughter is doing a college assignment and I managed to help her with the majority of the question but cannot figure out the last part.

I have the characteristic equation and have made that a cubic but unsure on the M^-1 part

Thanks - Apr 11th 2012, 01:22 AMDevenoRe: Help with question on matrices
ok, this is what i have, but you may want to check my arithmetic, because i DO make mistakes:

the Cayley-Hamilton theorem tells us M satisfies its own characteristic equation, which is $\displaystyle (x+1)(x-2)(x-4) = x^3 - 5x^2 + 2x + 8$.

therefore: $\displaystyle M^3 - 5M^2 + 2M + 8I = 0$, or put another way:

$\displaystyle M^3 = 5M^2 - 2M - 8I$.

the trouble is now how to express $\displaystyle M^2$ in terms of $\displaystyle M^{-1},M,I$. to accomplish that, we write:

$\displaystyle M^3 - 5M^2 + 2M = -8I$

$\displaystyle \frac{-1}{8}(M^2 - 5M + 2I)(M) = I$, so that evidently:

$\displaystyle M^{-1} = \frac{-1}{8}(M^2 - 5M + 2I)$, or:

$\displaystyle -8M^{-1} = M^2 - 5M + 2I$, and re-arranging:

$\displaystyle M^2 = 5M - 2I - 8M^{-1}$.

substituting back in our original equation:

$\displaystyle M^3 = 5M^2 - 2M - 8I = 5(5M - 2I - 8M^{-1}) - 2M - 8I$

$\displaystyle = 23M - 18I -40M^{-1}$, obtaining:

p = 23, q = -18, r = -40. - Apr 11th 2012, 05:23 AMJane1947Re: Help with question on matrices
Thanks very much for the time and effort you have put into this. I appreciate the answer and shall sit down with a pencil and try and explain it to her!