How do we get the second matrix from the first one in this example

In summary: 1 0 0 00 1 0 00 0 1 00 0 0 11/12 0 0 0-1/12 1 0 0-13/12 0 1 0-11/12 0 0 11 0 -1/12 -1/120 12 143/12 11/120 12 143/12 11/120 12 -143/12
  • #1
Madou
42
0

Homework Statement


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?



Homework Equations





The Attempt at a Solution



 
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  • #2
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
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:

[tex]\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}[/tex]
 
  • #4
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
Use the tex delimiters. Just click on the matrix in my post, you should get a popup with the source code.
 
  • #6
Could you please say where are the TEX delimeters?
 
  • #7
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
ok i think i see now how to type it
 
  • #9
[tex] 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}[/tex]
They are located this way in a block matrix :
AB
CAnd how can I get this with elementary row operations?

[tex] 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}[/tex]
So I have a new matrix

AB
C
 
  • #10
Forget about A and B. You can do the operations on C alone. Look at rows 7 and 8.
 
  • #11
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:
  • #12
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
but this way C is not a 'reflection'...
 
  • #14
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
thank you very much, cellotim
 
  • #16
Hi Madou! Welcome to PF! :smile:

Hi cellotim! (are we related?:wink:)

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

1. What is the process of obtaining the second matrix from the first one?

The process of obtaining the second matrix from the first one involves performing a specific mathematical operation on the elements of the first matrix. This operation could be addition, subtraction, multiplication, or division.

2. Can the second matrix be obtained by simply copying the first one?

No, the second matrix cannot be obtained by simply copying the first one. It requires a mathematical operation to be performed on the elements of the first matrix in order to obtain the second matrix.

3. Is there a specific method for determining which mathematical operation to use?

Yes, there are different methods for determining which mathematical operation to use in order to obtain the second matrix. This depends on the desired outcome and the properties of the first matrix.

4. Are there any special rules or restrictions when obtaining the second matrix?

Yes, there may be special rules or restrictions when obtaining the second matrix. For example, the dimensions of the first and second matrix must be compatible for the chosen mathematical operation to be performed successfully.

5. What are some common mistakes when obtaining the second matrix?

Some common mistakes when obtaining the second matrix include using the wrong mathematical operation, not considering the dimensions of the matrices, or making errors in the calculation process. It is important to double check all steps and calculations to ensure an accurate result.

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