Weird question

1. Oct 15, 2008

roam

Let $$A = \left(\begin{array}{ccc}3&0&0\\0&1&2\\0&0&1\end{array}\right)$$

Find a formula for An

3. The attempt at a solution

This is a weird question, I asked a few tutors for help but none of them were able to help me. They said that this is a weird question.

Does anyone here know how to do that? My friends & I can’t figure it out. Therefore I decided to post it here to see if anyone can show us how this is done.

We haven’t studied eigenvalues & Eigenvectors or diagonalization etc… we have only studied basic Linear Algebra, just as a warning!

I have three guesses;

(I) $$A^n = \left(\begin{array}{ccc}3&0&0\\0&1&2\\0&0&1\end{array}\right)^n$$

(II) $$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1^n&2^n\\0&0&1^n\end{array}\right)$$

(III) $$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1^n&0\\0&0&1^n\end{array}\right)$$+$$\left(\begin{array}{ccc}0&0&0\\0&0&2\\0&0&0\end{array}\right)$$

Is any of them close to the correct answer?

2. Oct 15, 2008

Defennder

(II) comes pretty close to the correct answer. You should that the diagonal entries of A^n are pretty easy to figure out by trying out A^n for small N. The only problem is the a23 entry. Again, just like above, try it out for small values of n. This follows an easy to identify pattern.

3. Oct 16, 2008

roam

I know that the a23 entry is a problem & that's why (in my third guess) I suggested to separate it from the rest.

So you are suggesting that I substitute small values for n ie. n = 1,2... & then An+1 etc ? Could you please explain what can we exactly conclude from that?

$$A^1 = \left(\begin{array}{ccc}3&0&0\\0&1&2\\0&0&1\end{ar ray}\right)$$

$$A^2 = \left(\begin{array}{ccc}9&0&0\\0&1&4\\0&0&1\end{ar ray}\right)$$

$$A^3 = \left(\begin{array}{ccc}27&0&0\\0&1&8\\0&0&1\end{ar ray}\right)$$

EDIT: The only pattern I see is that a11 & a23 are the two entries that actually increase by every n.

4. Oct 16, 2008

Defennder

Your A^3 is incorrect. Recheck the entries. They are increasing, but how exactly are they increasing? I did say that your original answer was nearly correct.

5. Oct 16, 2008

roam

They are increasing in such pattern;

$$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1&2^n\\0&0&1\end{array}\right)$$

$$A^2 = \left(\begin{array}{ccc}9&0&0\\0&1&4\\0&0&1\end{ar ray}\right)$$

$$A^{n+1} = A^{2+1} => A^3 = \left(\begin{array}{ccc}729&0&0\\0&1&64\\0&0&1\end{ar ray}\right)$$

6. Oct 16, 2008

roam

Is it something like:
If n = 2
A2 = 9
A2+1 = 92+1 = 729

Please tell me I'm so curious!

7. Oct 16, 2008

This is an exercise in 'block matrix' multiplication. let

$$A = \begin{bmatrix} 1 & \mathbf{0}'\\ \mathbf{0} & B\end{bmatrix}$$

where

$$\mathbf{0} = \begin{bmatrix} 0\\ 0 \end{bmatrix}, B = \begin{bmatrix} 1 & 2\\0 & 1 \end{bmatrix}$$

Here

$$A^n = \begin{bmatrix} 1 & \mathbf{0}' \\ \mathbf{0} & B^n \end{bmatrix}$$

8. Oct 16, 2008

HallsofIvy

Staff Emeritus
Using statdad's idea, let M= [3] and
$$N= \left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right]$$
so that
$$A= \left[\begin{array}{cc} M & 0 \\ 0 & N\end{array}\right]$$
and
$$A^n= \left[\begin{array}{cc}M^n & 0 \\ 0 & N^n\end{array}\right]$$
Mn is trivial and Nn should be pretty easy to see after calculating a few low powers. (Which is always a good idea in problems like these!)

9. Oct 16, 2008

HallsofIvy

Staff Emeritus
No, that's not right at all! What is
$$\left(\begin{array}{ccc}9 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right)$$

No! I am seriously tempted to ask why you are trying to do matrix arithmetic if you can't do arithmetic with numbers! A3 is NOT 93 because A is not 9. If A2 is 9, then A is either 3 or -3. If A= 3, then A3= 33= 27. If A= -3, A3= (-3)3= -27.

10. Oct 17, 2008

roam

Forgive my blur, I had a terrible headache along with a great hangover, and I appologize for the mistakes. Here's the corrected version of what I think:

.......[3 0 0]
A¹ = [0 1 2]
.......[0 0 1]

.......[3² 0 0]
A² = [0 1 4 ]
.......[0 0 1 ]

........[3³ 0 0]
A³ = [0 1 6 ]
........[0 0 1 ]

........[3⁴ 0 0]
A⁴ = [0 1 8 ]
........[0 0 1 ]
:
:
:
:
........[3ⁿ 0 0]
An = [0 1 2n ]
........[0 0 1 ]

$$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1&2n\\0&0&1\end{ar ray}\right)$$

Last edited: Oct 17, 2008
11. Oct 17, 2008

Tobias Funke

Here's another way to approach the problem and find the pattern. Have you learned that this matrix represents a certain linear transformation of R^3 to itself and what the columns of the matrix mean in that case? This transformation takes the vector (1,0,0) to (3,0,0), tripling its first coordinate. It takes (0,1,0) to (0,2,0), doubling its second coordinate, and it takes (0,0,1) to (0,2,1), adding 2 to its second coordinate.

Now if you know that matrix multiplication represents function composition, can you see what happens when you carry out this transformation n times? Looks like you do, but you said above that that's what you think, so if you're still unsure, can you prove it (by induction maybe)?

12. Oct 17, 2008

Defennder

That's complicating the problem unnecessarily. The OP hasn't learned eigvenvectos or diagonalisation, and he certainly doesn't even need to understand linear transformations to do this. He got it right with his last post, though.

13. Oct 18, 2008

Tobias Funke

Yeah I also misread the second column as (0,2,0). I don't know exactly what the OP knows, but you don't need to know anything about eigenvectors or diagonalization to know what I said. At least in my case, thinking in those terms actually makes the problem seem simpler and gives the pattern right away. But you're right of course-it's not necessary to solve the problem.