1. ## matrices problem

Question:
Given that $\displaystyle P=\begin{pmatrix} 3 &4 \\ -4 &3 \end{pmatrix}$ and $\displaystyle A = \begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix}$, find the martix M where $\displaystyle \mathbf{M=P^{-1}AP}$. Hence or otherwise find $\displaystyle \mathbf{M^{5}}$

Solution:

$\displaystyle \mathbf{M=P^{-1}AP}$
$\displaystyle M=\frac{1}{25} \begin{pmatrix} 3&-4 \\ 4& 3 \end{pmatrix} \begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix} \begin{pmatrix} 3 & 4\\ -4 & 3 \end{pmatrix}$
$\displaystyle M= \frac{1}{25} \begin{pmatrix} 3&-4 \\ 4& 3 \end{pmatrix} \begin{pmatrix} 6+0 &8+0 \\ 0+(-4) &0+3 \end{pmatrix}$
$\displaystyle M=\frac{1}{25} \begin{pmatrix} 3&-4 \\ 4& 3 \end{pmatrix} \begin{pmatrix} 6 &8 \\ -4&3 \end{pmatrix}$
$\displaystyle M= \frac{1}{25} \begin{pmatrix} 18+16&24-12 \\ 24-12& 32+9 \end{pmatrix}$
$\displaystyle \frac{1}{25} \begin{pmatrix} 34&12 \\ 12& 41 \end{pmatrix}$

My problem now is how do i find

$\displaystyle \mathbf{M^{5}}$
Am i suppose to raise each number in the matrix to the power 5 ( eg; $\displaystyle 34^5$) or raise the matrix M to the power 5

the answer given by the book is
$\displaystyle \frac{1}{25} \begin{pmatrix} 304&372 \\ 372& 521 \end{pmatrix}$

2. ## Re: matrices problem

The idea is that $M= P^{-1}AP$ where $A$ is diagonal.

$M^5 = P^{-1}AP \cdot P^{-1}AP \cdot \cdot \cdot P^{-1}AP = P^{-1}A^5 P$

$(A^5)_{i,i} = (A_{i,i})^5$

3. ## Re: matrices problem

Originally Posted by bigmansouf
Question:
Given that $\displaystyle P=\begin{pmatrix} 3 &4 \\ -4 &3 \end{pmatrix}$ and $\displaystyle A = \begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix}$, find the martix M where $\displaystyle \mathbf{M=P^{-1}AP}$. Hence or otherwise find $\displaystyle \mathbf{M^{5}}$
Here is the trick:
$\displaystyle \underbrace {{P^{ - 1}}AP \cdots {P^{ - 1}}AP}_{5 \times } = {P^{ - 1}}{A^5}P$

But $\displaystyle {\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}}\right )^5 ={\left( {\begin{array}{*{20}{c}}32&0\\0&1\end{array}} \right)}}$

4. ## Re: matrices problem

With matrices, $\displaystyle M^5$ means raise the matrix M to the fifth ie MxMxMxMxM.

For diagonal matrices (all non-diagonal elements are 0) such as A you can calculate $\displaystyle A^5$ by simply raising the diagonal elements to the fifth. I suggest you find $\displaystyle A^2$ by multiplying $\displaystyle A*A$ and it'll make sense why this is so for diagonal matrices.

As Romsek said
$\displaystyle M^5 = (P^{-1}AP) *(P^{-1}AP) *(P^{-1}AP) *(P^{-1}AP) *(P^{-1}AP) = P^{-1}*(AP *P^{-1})*(AP *P^{-1})*(AP *P^{-1})*(AP *P^{-1})*AP$ using the associativity of matrix multiplication
$\displaystyle = P^{-1}*A*A*A*A*A*P$
$\displaystyle = P^{-1}*A^5* P$

Now since $\displaystyle A^5$ is easy to calculate (since A is diagonal), it is easier to calculate $\displaystyle = P^{-1}*A^5* P$ than $\displaystyle M^5$.