4x4 matrix that satisfies conditions

1. Feb 9, 2013

burton95

A = [aij]

1) aij = i +j
2) aij = i^j-1
3) aij = 1 if |i - j| >1
-1 if |i - j| _< 1

I dont even know where to begin. Are i and j compenents of the matrix? Please help me get started

2. Feb 9, 2013

Dick

i and j and indices of the components. I would guess they take the values 1,2,3,4. So 1) would say a11=1+1=2, a12=1+2=3. Etc. Just write out the whole matrix in each case.

3. Feb 9, 2013

Dick

Mod note: I removed the copied text that Dick refers to, below.
The three different conditions 1), 2) and 3) describe different matrices. Which one are you doing? And judging by the title, it's supposed to be 4x4.

Last edited by a moderator: Feb 10, 2013
4. Feb 9, 2013

phion

Oops, you're right. The first 4x4 should go like...

$$A_{i,j} = \begin{pmatrix} a_{1+1} & a_{1+2} & \cdots & a_{1+j} \\ a_{2+1} & a_{2+2} & \cdots & a_{2+j} \\ \vdots & \vdots & \ddots & \vdots \\ a_{i+1} & a_{i+2} & \cdots & a_{i+j}\\ \end{pmatrix}$$

5. Feb 9, 2013

Dick

I admire your texing skills but that's pretty strange looking as an answer to the question. The matrix will have numerical entries. And besides, the goal here is not even to give answers. It's to show the poster how to solve it.

6. Feb 10, 2013

phion

I'm aware how the answer should look, and thank you for the compliment. I'm still trying to learn LaTeX, so I thought this would be an ample opportunity. I am only trying to help.

7. Feb 10, 2013

burton95

Thanks. I was trying to come up with one matrix to satisfy all the conditions mentioned. Upon review its states "condition". Wheeewh

8. Feb 10, 2013

burton95

so I got these as my matrices

1)
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8

2)
1 1 1 1
1 2 4 8
1 3 9 27
1 4 16 64

3)
-1 -1 -1 -1
-1 -1 -1 -1
1 -1 -1 -1
1 1 -1 -1

yes, no?

9. Feb 10, 2013

Dick

The first one looks ok. The second one is ok if the formula is i^(j-1). I'd read i^j-1 as (i^j)-1. For 3) shouldn't there be some ones in the upper right corner too?

10. Feb 10, 2013

burton95

you're correct on the 3rd matrix. The notation for 2) is aij = ij-1. I use the physicsforums.com android app and when I post using the app there is no template. Does this mean that using the app to post questions isn't legitimate? Also is there a way to keep score where I can thank folks for the help?

thanks

Last edited: Feb 10, 2013
11. Feb 10, 2013

Dick

It's odd the android app doesn't give you a template. But posting questions anyway you like is legit. Just show how you attempted to solve it before asking for help. And a simple thanks is fine. And you just did that. You are welcome!