Question

Need help with a question related to matrices. Our instructor didn't teach us how to do this but it's in our assignment as a compulsory question.
Let M= [-2 1, -36 10].
Find formulas for M^n where n is a positive integer.
M^n=[? ?, ? ?]

Answer 1

Here's the question as it's shown in the assignment.

Answer 2

Assume that we want to find \[M ^{1}, M^2, M^3, ...\]

Answer 3

... where

Answer 4

the first is just the matrix M itself, and \[M^2\]

... is the square of matrix M, and so on.

Answer 5

Take matrix M and multiply it by itself, of course following the rules for matrix multiplication. Try this, please, showing all work on which you'd like to receive feedbacl/

Answer 6

Okay, I got M^2 as [-32 8, -288 64]. Now what should I do?

Answer 7

I'm going to assume that your answer is correct.
To find M^3, multiply your most recent matrix result by M.

Answer 8

M^3=[-224 48, 1728 352]

Answer 9

Glad to say, we've both gotten the same result for M^2.
I won't be doing the entire problem myself, but encourage you to find M^3 and M^4.
Your next step is to look at the element in the first row, first column of M^1, M^2, M^3 and M^4 to determine whether that element seems to be predictable depending upon the exponent n.

Answer 10

If you have a TI-83 or -84 or more advanced, more recent calculator, you can do matrix multiplication with it. Strongly encourage you to learn how to do this.

Answer 11

Notice that this problem asks you for the contents of four different "answer boxes."
This means you need to try to find a formula for each such box, so that you can predict the numerical value of any such box if given 2x2 matrix M and exponent n.

Answer 12

\[\textbf M= \begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\]
\[\textbf M^2=\textbf M\textbf M= \begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\]
\[\textbf M^3=\textbf M\textbf M\textbf M= \begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\]

Answer 13

you have \(\textbf M ^2\) correct

Answer 14

\[\textbf M^2=\textbf M\textbf M= \begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}
=\begin{bmatrix}-32 &8\\ -288& 64\end{bmatrix}\]

\[\textbf M^3=\textbf M\textbf M^2= \begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\begin{bmatrix}-32 &8\\ -288& 64\end{bmatrix}\]

Answer 15

Unkle: Your input is helpful and surely looks good. All I ask is that you not do for the other person what that person could do for himself or herself. Thanks.

Answer 16

Arisa.S already has shown the correct answers for M^2,^3

Answer 17

except there is a sign error in M^3

Answer 18

Unkle: Thanks for noticing that. Perhaps you could engage Arisa in locating and fixing the sign error.

Answer 19

it's supposed to be -1728 in the second row first column, right?

Answer 20

that's it

Answer 21

So you have
\[\textbf M= \begin{bmatrix}-2 &1\\ -36 &10\end{bmatrix}\]\[\textbf M^2=\begin{bmatrix}-32 &8\\ -288& 64\end{bmatrix}\]\[\textbf M^3=\begin{bmatrix}-244&48\\-1728&352\end{bmatrix}\]

Answer 22

hmm, this doesn't resolve into a nice pattern, i think this is a dead end.
We might need a different method

Answer 23

Can you diagonalize the matrix \(\mathbf M\)?

Answer 24

Agreed...no distinct pattern has hit me yet. Pls explain to Arisa what you mean by "diagonalize the matrix M," and again to join our dialogue.

Answer 25

I couldn't figure out a relation either.

Answer 26

Diagonalisation is the process of finding a matrix \(\mathbf{P}\) such that
\[
\mathbf{A}=\mathbf{PD}\mathbf{P}^{-1}\text{ or }\mathbf{D}=\mathbf{P^{-1}}\mathbf{AP}
\]
\(\mathbf{D}\) of course is a diagonal matrix. If \(\mathbf{A}=\mathbf{PD}\mathbf{P}^{-1}\), then \(\mathbf{A}^n=\mathbf{PD}^n\mathbf{P^{-1}}\).

The only problem is that this matrix is not diagonalisable. Have your teacher taught you the Jordan normal form yet?

Answer 27

I used the computer software Mathematica to generate a closed-form solution but I don't see how you can get it without using Jordan normal form.