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 December 29th, 2015, 03:41 AM #1 Newbie   Joined: Dec 2015 From: Karachi Posts: 2 Thanks: 0 Column Matrix Expansion Ok guys I want to ask a question that if we open a matrix by column then should we rotate the minor ? example: |1 1 -1 | |2 0 1 | |2 -1 -3 | I expand from C2 1|2 1| -0 +(-1)|1 -1| = (-6-2)-(1+2) = -11 |2 -3| |2 1| Now I expand from R1 1| 0 1| -1|2 1| +(-1)|2 0| = (0+1)-(-6-2)-(-2-0)= 1+8+2=11 |-1 -3| |2 -3| |2 -1| Determinant is different for both expansions Now I make some difference Again taking C2 1|1 -3| -0 +(-1)|-1 1| = (2+6)-(-2-1) =8+3= 11 |2 2| |1 2| This time I rotated the 3x3 matrix before writing 2x2 determinants down Now you can see that before rotating the answer with row expansion did not match but after rotating it is matching But this thing seems new to me because once before my teacher didn't tell me anything about this even when I passed XI grade, now I am repeating it in some other city and my teacher told me this new thing Question: Is the rotating method an official method or it doesn't even exist ? Last edited by Ogie; December 29th, 2015 at 03:43 AM.
December 29th, 2015, 07:05 AM   #2
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Quote:
 Originally Posted by Ogie Ok guys I want to ask a question that if we open a matrix by column then should we rotate the minor ? example: |1 1 -1 | |2 0 1 | |2 -1 -3 | I expand from C2 1|2 1| -0 +(-1)|1 -1| = (-6-2)-(1+2) = -11 |2 -3| |2 1|
You have expanded wrong. You do NOT just start with a random number and then alternate sign. Each term is multiplied by $(-1)^{i+j}$ where i and j are the row and column. Since you are expanding on the middle column, for the first number i= 1 and j= 2 so i+ j= 3. The first term is multiplied by [tex](-1)^{1+ 2}= (-1)^3= -1. The next two terms have j= 2 but i= 2 then 3 so i+j= 4 and 5. You should have -1(1)(-6- 2)+ 1(0)- 1(-1)(1+ 2)= 8+ 3= 11, not -11.

Quote:
 Now I expand from R1
Now, i= 1 while j= 1, 2, 3 so i+j= 2, 3, 4 and the signs are 1, -1, 1 as you have.
Quote:
 1| 0 1| -1|2 1| +(-1)|2 0| = (0+1)-(-6-2)-(-2-0)= 1+8+2=11 |-1 -3| |2 -3| |2 -1| Determinant is different for both expansions Now I make some difference Again taking C2 1|1 -3| -0 +(-1)|-1 1| = (2+6)-(-2-1) =8+3= 11 |2 2| |1 2| This time I rotated the 3x3 matrix before writing 2x2 determinants down Now you can see that before rotating the answer with row expansion did not match but after rotating it is matching But this thing seems new to me because once before my teacher didn't tell me anything about this even when I passed XI grade, now I am repeating it in some other city and my teacher told me this new thing Question: Is the rotating method an official method or it doesn't even exist ?
No, you are simply mistaken about how the signs work in expanding a determinant.

 December 29th, 2015, 10:56 AM #3 Newbie   Joined: Dec 2015 From: Karachi Posts: 2 Thanks: 0 Thank you You solved my problem But I still have a question, why was the determinant of 3x3 matrix same upon rotating the 2x2 determinant ?
December 31st, 2015, 10:23 AM   #4
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Quote:
 Originally Posted by Ogie But I still have a question, why was the determinant of 3x3 matrix same upon rotating the 2x2 determinant ?
Rotating a 2X2 matrix switches the sign of its determinant.

if A = ( a b ) , then detA = ad - bc
( c d )

Set B to be the counter-clockwise rotation of A.

B = ( b d ) detB = bc - ad = -detA
( a c )

Because your original mistake only caused a switch of each term's sign, rotation of each term's minor matrix corrected the mistake by switching it back.

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