# Homework Help: How do we get the second matrix from the first one in this example

1. Oct 3, 2008

1. The problem statement, all variables and given/known data
I have a matrix 1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
0, 12, 143/12, 11/12 |-13/12,0,1,0
0, -12, -143/12, -11/12|-11/12,0,0,1
---------------------------------------------------
1 , 0, 0 , 0 |
0, 1, 0 , 0 |
0, 0, 1, 0 |
0, 0, 0 , 1 |
And I can't understand how can one get the following block matrix from the previous one with elementary row operations:
1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
0, 12, 143/12, 11/12 |-13/12,0,1,0
0, -12, -143/12, -11/12|-11/12,0,0,1
---------------------------------------------------
1 , 0, -1/12 ,-1/12 |
0, 1, 0 , 0 |
0, 0, 1, 0 |
0, 0, 0 , 1 |

And ideas?

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 3, 2008

I have a matrix:
1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
0, 12, 143/12, 11/12 |-13/12,0,1,0
0, -12, -143/12, -11/12|-11/12,0,0,1
---------------------------------------------------
1 , 0, 0 , 0 |
0, 1, 0 , 0 |
0, 0, 1, 0 |
0, 0, 0 , 1 |
And I can't understand how can one get the following block matrix from the previous one with elementary row operations:
1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
0, 12, 143/12, 11/12 |-13/12,0,1,0
0, -12, -143/12, -11/12|-11/12,0,0,1
---------------------------------------------------
1 , 0, -1/12 ,-1/12 |
0, 1, 0 , 0 |
0, 0, 1, 0 |
0, 0, 0 , 1 |

And ideas?

3. Oct 3, 2008

### cellotim

That's a bit confusing. I see what looks like three matrices. Can you rewrite it using the latex array format? Example: \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array} gives:

$$\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}$$

4. Oct 3, 2008

And how do I put this ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}) into a message (reply to thread)?

5. Oct 3, 2008

### cellotim

Use the tex delimiters. Just click on the matrix in my post, you should get a popup with the source code.

6. Oct 3, 2008

Could you please say where are the TEX delimeters?

7. Oct 3, 2008

### cellotim

You type them [x] \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array} [/x] where x=tex. Use cccc for a 4x4 matrix instead of ccc.

8. Oct 3, 2008

ok i think i see now how to type it

9. Oct 3, 2008

$$Matrix A: \begin{array}{cccc} 1 & 0 & 1/12 & 1/12 \\ 0 & 12 & 143/12 & 11/12\\ 0 & 12& 143/12 & 11/12 \\ 0 & -12 & -143/12 & -11/12\end{array} Matrix B : \begin{array}{cccc} 1/12 & 0 & 0 & 0 \\ -1/12 & 1 & 0 & 0\\ -13/12 & 0& 1 & 0 \\ -11/12 & 0 & 0 & 1\end{array} Matrix C : \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0& 1 & 0 \\ 0 & 0 & 0 & 1\end{array}$$
They are located this way in a block matrix :
AB
C

And how can I get this with elementary row operations?

$$Matrix A: \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 12 & 143/12 & 11/12\\ 0 & 12& 143/12 & 11/12 \\ 0 & -12 & -143/12 & -11/12\end{array} Matrix B : \begin{array}{cccc} 1/12 & 0 & 0 & 0 \\ -1/12 & 1 & 0 & 0\\ -13/12 & 0& 1 & 0 \\ -11/12 & 0 & 0 & 1\end{array} Matrix C : \begin{array}{cccc} 1 & 0 &-1/12 & -1/12 \\ 0 & 1 & 0 & 0\\ 0 & 0& 1 & 0 \\ 0 & 0 & 0 & 1\end{array}$$
So I have a new matrix

AB
C

10. Oct 3, 2008

### cellotim

Forget about A and B. You can do the operations on C alone. Look at rows 7 and 8.

11. Oct 3, 2008

hmm... I see what you mean. But the thing is I thought in this example I was doing things with matrix A and matrices B and C were the 'reflections' of what I was doing with matrix A.
And what about the first row in matrix A? You see it has changed also.

Last edited: Oct 3, 2008
12. Oct 3, 2008

ohh, I think I see now the answer to my question.

I can now add the 7th row multiplyed by -1/12 and the 8th multiplyed by -1/12 to the first one and I GET WHAT I SEE! ;)

13. Oct 3, 2008

but this way C is not a 'reflection'...

14. Oct 4, 2008

### cellotim

I see. In your original question, A's first row was unchanged. In that case, just use the new C row to get the result.

15. Oct 4, 2008

thank you very much, cellotim

16. Oct 9, 2008

### tiny-tim

Hi cellotim! (are we related?)

There's an easier way to do matrices: use the CODE tag (next to the QUOTE tag)…

it's really intended for writing computer code, but it does matrices nicely, because it prints with equal spacing, as on a typewriter …

Code (Text):

1    0    1/12   1/12
0   12  143/12  11/12
0   12  143/12  11/12
0   12 -143/12 -11/12