Find C44 for the 4x4 matrix A given: Cofactor of A = -3

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The discussion revolves around calculating the cofactor C44 of a given 4x4 matrix A, with an initial cofactor value of -3. Participants engage in a detailed step-by-step approach to derive C44, correcting each other's mistakes along the way. The final calculation leads to the conclusion that C44 equals -166, after addressing errors in previous calculations and clarifying the rules for manipulating matrix rows. The importance of using the correct method to simplify the matrix for easier determinant calculation is emphasized. Ultimately, the correct value of C44 is confirmed as -166.
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For the matrix A given below, find C44 , where Cij is the i,j cofactor of A.

A=

| -5 +4 +5 -3 |
| +2 -1 +5 +4 |
| -5 -1 +2 -3 |
| +4 -2 +4 -4 |

So this is how I answer this question:

= (4,4) cofactor of A

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| +5 +1 -2 |
R3*(-1)

=
| +0 +5 +3 |
| +2 -1 +5 |
| +5 +1 -2 |
Add R1 and R3

=
| -1 +5 | *(+)(+)
| +1 -2 |
(1,1) cofactor of A

=(+)(+)[(-1*-2)-(1*5)]

=(+)(+)(2-5)

=-3

Please help me on this, is C44 = -3?
 
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Hi shiri! :smile:
shiri said:
… =
| +0 +5 +3 |
| +2 -1 +5 |
| +5 +1 -2 |
Add R1 and R3

=
| -1 +5 | *(+)(+)
| +1 -2 |
(1,1) cofactor of A

=(+)(+)[(-1*-2)-(1*5)]

No, that's the cofactor of the +0, and you get the determinant by multiplying +0 times its cofactor (and then adding the same for +5 and +3).

If you're determined to save effort by getting down to a 2x2 determinant, you need another 0.

I'd have started differently, and used one of the original -1s to get rid of the other -1 and the 4. :wink:
 
tiny-tim said:
Hi shiri! :smile:

… =
| +0 +5 +3 |
| +2 -1 +5 |
| +5 +1 -2 |
Add R1 and R3

=
| -1 +5 | *(+)(+)
| +1 -2 |
(1,1) cofactor of A

=(+)(+)[(-1*-2)-(1*5)]

No, that's the cofactor of the +0, and you get the determinant by multiplying +0 times its cofactor (and then adding the same for +5 and +3).

If you're determined to save effort by getting down to a 2x2 determinant, you need another 0.

I'd have started differently, and used one of the original -1s to get rid of the other -1 and the 4. :wink:

Hi

… =
| +0 +5 +3 |
| +2 -1 +5 |*(+)
| +5 +1 -2 |
Add R1 and R3

=
| +0 +5 +3 |
| +2 -1 +5 |*(+)
| +0 +7 -29 |
2R3 - 5R2

=
| +5 +3 | *(+)(-)
| +7 -29 |
(2,1) cofactor of A

=(+)(-)[(+5*-29)-(+7*+3)]

=(+)(-)[(-145)-(21)]

C44 = 166

So after reading your message, is this a correct answer for C44?
 
hmm … you've made two mistakes, and still got the right answer! :rolleyes: :biggrin:

i] you can add as many copies as you like of one row to another row,

but you can't add copies of a row to itself … if you do that, you multiply the determinant (in this case by 2, when you added R3 - 5R2 to the original R3)

ii] you should have multiplied the cofactor, 166, by the 2 that it's the cofactor of.

Try again. :smile:
 
tiny-tim said:
hmm … you've made two mistakes, and still got the right answer! :rolleyes: :biggrin:

i] you can add as many copies as you like of one row to another row,

but you can't add copies of a row to itself … if you do that, you multiply the determinant (in this case by 2, when you added R3 - 5R2 to the original R3)

ii] you should have multiplied the cofactor, 166, by the 2 that it's the cofactor of.

Try again. :smile:

So the C44 = 332?
 
shiri said:
So the C44 = 332?

No!

I said 166 was right …

you made two mistakes, one doubled the answer, and the other halved it again.

Do you understand why your 2R3 - 5R2 was wrong?​
 
tiny-tim said:
No!

I said 166 was right …

you made two mistakes, one doubled the answer, and the other halved it again.

Do you understand why your 2R3 - 5R2 was wrong?​

Strange. When I typed 166 in the online assignment. It came out wrong. Do you think the answer was suppose to be -166, not 166?
 
(just got up :zzz: …)
shiri said:
Strange. When I typed 166 in the online assignment. It came out wrong. Do you think the answer was suppose to be -166, not 166?

oops! yes, I didn't notice that you made a third mistake (I thought that the two matrices were the same :rolleyes:) …
shiri said:
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| +5 +1 -2 |
R3*(-1)

when you multiply a row by minus one, you multiply the whole determinant by minus one!

As I said before, you can only add multiples of one row to another row! :smile:

Now try the easier way I suggested before: use one of the original -1s to get rid of the other -1 and the 4. :wink:
 
tiny-tim said:
(just got up :zzz: …)


oops! yes, I didn't notice that you made a third mistake (I thought that the two matrices were the same :rolleyes:) …


when you multiply a row by minus one, you multiply the whole determinant by minus one!

As I said before, you can only add multiples of one row to another row! :smile:

Now try the easier way I suggested before: use one of the original -1s to get rid of the other -1 and the 4. :wink:


So it becomes like this:

| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +5 | * (+) (+)
| -5 +2 |

=(+)(+)[(-5*+2)-(-5*+5)]

=(+)(+)[-10-(-25)]

=(+)(+)[-10+25]

=(+)(+)[15]

So is that means, C44 = 15?
 
  • #10
That's bizarre! :redface:

All you've done is calculate the cofactor of (an arbitrary) -1.

Add multiples of the 2nd row to the 1st and 3rd rows so as to turn the -1 and the 4 into 0s …

then! you can use the cofactor of the remaining -1 ! :smile:
 
  • #11
tiny-tim said:
That's bizarre! :redface:

All you've done is calculate the cofactor of (an arbitrary) -1.

Add multiples of the 2nd row to the 1st and 3rd rows so as to turn the -1 and the 4 into 0s …

then! you can use the cofactor of the remaining -1 ! :smile:

OK. I hope I know what I'm doing.

| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| +7 0 +3 |
R3(-1) + R2

=
| +3 +0 +25 |
| +2 -1 +5 | * (+)
| +7 +0 +3 |
R1 + 4R2

=
| +3 +25 | *(+)(+)
| +7 +3 |

=(+)(+)[(+3*+3)-(+7*+25)

=(+)(+)[+9-175]

=-166

C44 = -166

Am I doing this right, tiny-tim?
 
  • #12
Yes! :smile:

Do you see why this works? …

the determinant is the sum of each element in one row (or one column) multiplied by its cofactor …

so you made a row that was all 0s except for one element …

so the determinant is that element multiplied by its cofactor!

(and you chose -1 because keeping the numbers small, and preferably ±1, makes the arithmetic a lot easier than eg 2 and 5 :wink:)
 
  • #13
tiny-tim said:
Yes! :smile:

Do you see why this works? …

the determinant is the sum of each element in one row (or one column) multiplied by its cofactor …

so you made a row that was all 0s except for one element …

so the determinant is that element multiplied by its cofactor!

(and you chose -1 because keeping the numbers small, and preferably ±1, makes the arithmetic a lot easier than eg 2 and 5 :wink:)

Great. Thank You! You're very helpful.

Oh, one more thing. Can you take a look at this question [https://www.physicsforums.com/showthread.php?t=346994 Matrix Determinant]

I just want to know whether I answer this question correctly. Thanks
 

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