Hi I can find M^2 on my calculator but what is the hand written method?

Thanks

M=[[1,-2][-2,3]]

My calculator says m^2=[[5,-8][-8,13]]

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- May 30th 2010, 10:25 PM #1

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- May 30th 2010, 10:40 PM #2
I assume that you mean

$\displaystyle M = \left[\begin{matrix}\phantom{-}1 & -2\\ -2 & \phantom{-}3\end{matrix}\right]$.

If so, then

$\displaystyle M^2 = MM$

$\displaystyle = \left[\begin{matrix}\phantom{-}1 & -2\\ -2 & \phantom{-}3\end{matrix}\right]\left[\begin{matrix}\phantom{-}1 & -2\\ -2 & \phantom{-}3\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix}(1)(1) + (-2)(-2) & (1)(-2) + (-2)(3) \\ (1)(-2) + (-2)(3) & (-2)(-2) + (3)(3)\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix}\phantom{-}1 + 4 & -2 - 6\\ -2 - 6 & \phantom{-}4 + 9\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix}\phantom{-}5 & -8\\ -8 & \phantom{-}13\end{matrix}\right]$

- May 30th 2010, 10:46 PM #3
$\displaystyle \begin{bmatrix}1 & -2\\ -2 & 3\end{bmatrix}\begin{bmatrix}1 & -2\\ -2 & 3\end{bmatrix}=?$

What we want to do is perform a dot product type of calculation to calculate each entry in the resulting matrix. We want to take the first row of the first matrix transposed and dot it with the first column of the second matrix:

$\displaystyle \begin{bmatrix}{\color{red}1} & {\color{red}-2}\\ -2 & 3\end{bmatrix}\begin{bmatrix}{\color{blue}1} & -2\\ {\color{blue}-2} & 3\end{bmatrix}=\begin{bmatrix}({\color{red}1})({\c olor{blue}1})+({\color{red}-2})({\color{blue}-2}) & ?\\ ? & ?\end{bmatrix}=\begin{bmatrix}5 & ? \\ ? & ?\end{bmatrix}$

We do a similar thing to find the second element in the first row:

$\displaystyle \begin{bmatrix}{\color{red}1} & {\color{red}-2}\\ -2 & 3\end{bmatrix}\begin{bmatrix}1&{\color{blue}-2}\\ -2 & {\color{blue}3} \end{bmatrix}=\begin{bmatrix}5 & ({\color{red}1})({\color{blue}-2})+({\color{red}-2})({\color{blue}3})\\ ? & ?\end{bmatrix}=\begin{bmatrix}5 & -8 \\ ? & ?\end{bmatrix}$

Can you follow this process and finish this multiplication?

- May 30th 2010, 10:57 PM #4

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- May 30th 2010, 11:43 PM #5
You can use the Cayley-Hamilton Theorem.

Cayley?Hamilton theorem - Wikipedia, the free encyclopedia

- May 31st 2010, 05:32 AM #6

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It took me a moment to figure out what Prove It meant!

The "Cayley-Hamilton Theorem" says that every matrix satisfies its own characteristic equation. For this particular matrix, the characteristic equation is $\displaystyle \left|\begin{array}{cc}1-\lambda & -2 \\ -2 & 3- \lambda\end{array}\right|= (1- \lambda)(3- \lambda)- 4= \lambda^2- 4\lambda- 1= 0$.

By the Cayley-Hamilton Theorem, then, $\displaystyle A^2= 4A- I$ and you can create higher powers from that: $\displaystyle A^6= (4A- I)^3$ which cuts the work by half.

It is also true that this matrix has two distinct eigenvalues, $\displaystyle 2+\sqrt{5}$ and $\displaystyle 2- \sqrt{5}$ and so have two independent eigenvectors. If we create the matrix P having two independent eigenvectors as columns, $\displaystyle A= PDP^{-1}$ where D is the diagonal matrix having the eigenvalues on its main diagonal: $\displaystyle \begin{bmatrix}2+ \sqrt{5} & 0 \\ 0 & 2- \sqrt{5}\end{bmatrix}$.

Then, for all n, $\displaystyle A^n= PD^nP^{-1}$ and powers of the diagonal matrix is just the diagonal matrix with powers of its original diagonal elements: $\displaystyle \begin{bmatrix}(2+ \sqrt{5})^n & 0 \\ 0 & (2- \sqrt{5})^n\end{bmatrix}$.