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anonymous
 2 years ago
A is a 2 × 2 matrix with eigenvectors
v1=[1;1] v2=[1;1] corresponding to eigenvalues
λ1 = (1/2) and λ2 = 2,respectively, and
X =[5;1]. Find A^10X.
anonymous
 2 years ago
A is a 2 × 2 matrix with eigenvectors v1=[1;1] v2=[1;1] corresponding to eigenvalues λ1 = (1/2) and λ2 = 2,respectively, and X =[5;1]. Find A^10X.

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jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3hint: try to write X as a linear combination of v1 and v2

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0how do i determine that

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3ok, first of all have you learned what a linear combination is?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0not quite im very confused with this topic

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3ok. A linear combination is just the addition and scalar multiplication of two vectors. Take for example \[2\left(\begin{matrix}1 \\ 0\end{matrix}\right) + 3\left(\begin{matrix}0 \\ 1\end{matrix}\right)\] this is a linear combination of (1 0) and (0 1) because we are multiplying by scalars and adding vectors... does that make sense?

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3excellent! that is right.

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3OK, so now you are probably thinking... now what? :)

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3Well, let's write A^10 X using our new expression for X as a linear combination. We get \[A^{10}X = A^{10}(2v_1 + 3v_2) = A^{10}(2v_1) + A^{10}(3v_2)\] by distributing. Do you see that?

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3Cool! Now, we get \[2A^{10}v_1 + 3A^{10}v_2\] since scalars can be moved around.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so then we can get 2(\[\lambda ^{10}\]v1+3(\[\lambda _{2^{10}}\]v2

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i just dont know how to get the final answer

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3OK, well I personally wouldn't actually calculate a number to the 10th power, just leave it with an exponent. I'll type it below.

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3Maybe something like this \[\left(\begin{matrix}2^{9} + 3 \times 2^{10} \\ 2^{9} + 3 \times 2^{10}\end{matrix}\right)\]

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3After I've just added to the two vector together after multiplying by the scalars.

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3That 2^(9) came from \[2(\frac{1}{2})^{10} = 2 \times 2^{10} = 2^{9}\]

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3but however you want to write it :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok i tired that final answer and i got it incorrect

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3could you take a screenshot maybe of what the input looks like?

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3or else you could just input the decimal answer... maybe the computer would like that better...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i got it correct thanks so much for your help

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.3no problem, good luck on eigenvectors! :)

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0since u have the eigenvectors and eigenvalues of the matrix A then A can be "factored" as \[\large A=S^{1}\Lambda S \] where \[\large S=\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}\qquad\text{and}\qquad \Lambda=\begin{pmatrix} 1/2 & 0\\ 0 & 2 \end{pmatrix} \]

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0therefore \[\large A^{10}x=(S^{1}\Lambda S)^{10}x=(S^{1}\Lambda^{10}S)x \]
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