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Where did I make a mistake in expanding the determinant?
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I want to work on the left determinant to find the last one. I did as follows but, the machinery way resulted it $(+2/b)$ while I got $(-2/b)$ as coefficient. May I ask you what is my probable mistake? Thank you!
$$beginvmatrix
b+a&1&0 \ -b-1&-1&-2a \-a-b&-1&2b
endvmatrix=(1/b)beginvmatrix
b+a&b&0 \ -b-1&-b&-2a \-a-b&-b&2b
endvmatrix=(-2/b)beginvmatrix
b+a&b&0 \ b+1&+b&a \-a-b&-b&b
endvmatrix=$$
$$(-2/b)beginvmatrix
-a&b&0 \ -1&+b&a \a&-b&b
endvmatrix$$‎
linear-algebra matrices
asked Sep 8 at 11:22
B.B.
1467
add a comment |Â
up vote
3
down vote
favorite
I want to work on the left determinant to find the last one. I did as follows but, the machinery way resulted it $(+2/b)$ while I got $(-2/b)$ as coefficient. May I ask you what is my probable mistake? Thank you!
$$beginvmatrix
b+a&1&0 \ -b-1&-1&-2a \-a-b&-1&2b
endvmatrix=(1/b)beginvmatrix
b+a&b&0 \ -b-1&-b&-2a \-a-b&-b&2b
endvmatrix=(-2/b)beginvmatrix
b+a&b&0 \ b+1&+b&a \-a-b&-b&b
endvmatrix=$$
$$(-2/b)beginvmatrix
-a&b&0 \ -1&+b&a \a&-b&b
endvmatrix$$‎
linear-algebra matrices
asked Sep 8 at 11:22
B.B.
1467
1
In final step subtract 2nd column from 1st and write that. Otherwise sign will change.
– Prince Kumar
Sep 8 at 11:26
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I want to work on the left determinant to find the last one. I did as follows but, the machinery way resulted it $(+2/b)$ while I got $(-2/b)$ as coefficient. May I ask you what is my probable mistake? Thank you!
$$beginvmatrix
b+a&1&0 \ -b-1&-1&-2a \-a-b&-1&2b
endvmatrix=(1/b)beginvmatrix
b+a&b&0 \ -b-1&-b&-2a \-a-b&-b&2b
endvmatrix=(-2/b)beginvmatrix
b+a&b&0 \ b+1&+b&a \-a-b&-b&b
endvmatrix=$$
$$(-2/b)beginvmatrix
-a&b&0 \ -1&+b&a \a&-b&b
endvmatrix$$‎
linear-algebra matrices
asked Sep 8 at 11:22
B.B.
1467
I want to work on the left determinant to find the last one. I did as follows but, the machinery way resulted it $(+2/b)$ while I got $(-2/b)$ as coefficient. May I ask you what is my probable mistake? Thank you!
$$beginvmatrix
b+a&1&0 \ -b-1&-1&-2a \-a-b&-1&2b
endvmatrix=(1/b)beginvmatrix
b+a&b&0 \ -b-1&-b&-2a \-a-b&-b&2b
endvmatrix=(-2/b)beginvmatrix
b+a&b&0 \ b+1&+b&a \-a-b&-b&b
endvmatrix=$$
$$(-2/b)beginvmatrix
-a&b&0 \ -1&+b&a \a&-b&b
endvmatrix$$‎
linear-algebra matrices
asked Sep 8 at 11:22
B.B.
1467
asked Sep 8 at 11:22
B.B.
1467
asked Sep 8 at 11:22
B.B.
1467
asked Sep 8 at 11:22
B.B.
1467
1467
1
In final step subtract 2nd column from 1st and write that. Otherwise sign will change.
– Prince Kumar
Sep 8 at 11:26
add a comment |Â
1
In final step subtract 2nd column from 1st and write that. Otherwise sign will change.
– Prince Kumar
Sep 8 at 11:26
1
1
In final step subtract 2nd column from 1st and write that. Otherwise sign will change.
– Prince Kumar
Sep 8 at 11:26
In final step subtract 2nd column from 1st and write that. Otherwise sign will change.
– Prince Kumar
Sep 8 at 11:26
add a comment |Â
2 Answers
2
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up vote
4
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accepted
At the end you should multiply again by $-1$ taking $2/b$ in front.
answered Sep 8 at 11:23
gimusi
71.4k73786
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
Thanks for your time!
– B.B.
Sep 8 at 11:42
You are welcome! Bye
– gimusi
Sep 8 at 11:47
add a comment |Â
up vote
3
down vote
In the last passage you summed the second column to the first one and then multiplied the first one by $-1$ (or, in other words, you substituted $C_1to -C_1-C_2$) but you did not update the coefficient accordingly.
answered Sep 8 at 11:28
Saucy O'Path
3,671424
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
At the end you should multiply again by $-1$ taking $2/b$ in front.
answered Sep 8 at 11:23
gimusi
71.4k73786
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
Thanks for your time!
– B.B.
Sep 8 at 11:42
You are welcome! Bye
– gimusi
Sep 8 at 11:47
add a comment |Â
up vote
4
down vote
accepted
At the end you should multiply again by $-1$ taking $2/b$ in front.
answered Sep 8 at 11:23
gimusi
71.4k73786
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
Thanks for your time!
– B.B.
Sep 8 at 11:42
You are welcome! Bye
– gimusi
Sep 8 at 11:47
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
At the end you should multiply again by $-1$ taking $2/b$ in front.
answered Sep 8 at 11:23
gimusi
71.4k73786
At the end you should multiply again by $-1$ taking $2/b$ in front.
answered Sep 8 at 11:23
gimusi
71.4k73786
answered Sep 8 at 11:23
gimusi
71.4k73786
answered Sep 8 at 11:23
gimusi
71.4k73786
answered Sep 8 at 11:23
gimusi
71.4k73786
71.4k73786
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
Thanks for your time!
– B.B.
Sep 8 at 11:42
You are welcome! Bye
– gimusi
Sep 8 at 11:47
add a comment |Â
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
Thanks for your time!
– B.B.
Sep 8 at 11:42
You are welcome! Bye
– gimusi
Sep 8 at 11:47
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
Oh. Indeed, at the end, I did $-C_1+C_2$ and put it in $C_1$. Wasn't it an elementary operation so the sign would not be changed?
– B.B.
Sep 8 at 11:37
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
No it change since you are multiplying C1 by -1.
– gimusi
Sep 8 at 11:40
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
You can consider also that the operation is equivalent to C1-C2 in column 1 and then reversing the sign of C1.
– gimusi
Sep 8 at 11:41
Thanks for your time!
– B.B.
Sep 8 at 11:42
Thanks for your time!
– B.B.
Sep 8 at 11:42
You are welcome! Bye
– gimusi
Sep 8 at 11:47
You are welcome! Bye
– gimusi
Sep 8 at 11:47
add a comment |Â
up vote
3
down vote
In the last passage you summed the second column to the first one and then multiplied the first one by $-1$ (or, in other words, you substituted $C_1to -C_1-C_2$) but you did not update the coefficient accordingly.
answered Sep 8 at 11:28
Saucy O'Path
3,671424
add a comment |Â
up vote
3
down vote
In the last passage you summed the second column to the first one and then multiplied the first one by $-1$ (or, in other words, you substituted $C_1to -C_1-C_2$) but you did not update the coefficient accordingly.
answered Sep 8 at 11:28
Saucy O'Path
3,671424
add a comment |Â
up vote
3
down vote
up vote
3
down vote
In the last passage you summed the second column to the first one and then multiplied the first one by $-1$ (or, in other words, you substituted $C_1to -C_1-C_2$) but you did not update the coefficient accordingly.
answered Sep 8 at 11:28
Saucy O'Path
3,671424
In the last passage you summed the second column to the first one and then multiplied the first one by $-1$ (or, in other words, you substituted $C_1to -C_1-C_2$) but you did not update the coefficient accordingly.
answered Sep 8 at 11:28
Saucy O'Path
3,671424
answered Sep 8 at 11:28
Saucy O'Path
3,671424
answered Sep 8 at 11:28
Saucy O'Path
3,671424
answered Sep 8 at 11:28
Saucy O'Path
3,671424
3,671424
add a comment |Â
add a comment |Â
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1
In final step subtract 2nd column from 1st and write that. Otherwise sign will change.
– Prince Kumar
Sep 8 at 11:26