Mixing
matrices, determinant [duplicate]
$begingroup$
This question already has an answer here:
Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)
9 answers
Determinant of a matrix will all off-diagonal elements equal
1 answer
i have the following homework problem:
Find the determinant of X(s):
X(s) =
[s, 1, 1, 1],
[1, s, 1, 1],
[1, 1, s, 1],
[1, 1, 1, s]
I know i can exploit the fact that the product of the diagonal gives me the determinant if the matrix is upper, lower or both triangular- but i don't know how i can turn it into a triangular?
If i try regular row operations, i get:
X(s) =
[s, 1, 1, 1],
[0, s-1, 1, 1],
[0, 0, s-1, 1],
[1-s, 0, 0, s-1]
This is by first subtracting R1 from R2-4, and then subtracting R4 from R2-3. But i can't get any further, how do i solve this?
linear-algebra determinant
asked Jan 30 at 19:50
melbilmelbil
31
$endgroup$
marked as duplicate by Dietrich Burde, Misha Lavrov, GNUSupporter 8964民主女神 地下教會, Cesareo, YiFan Feb 10 at 1:20
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)
9 answers
Determinant of a matrix will all off-diagonal elements equal
1 answer
i have the following homework problem:
Find the determinant of X(s):
X(s) =
[s, 1, 1, 1],
[1, s, 1, 1],
[1, 1, s, 1],
[1, 1, 1, s]
I know i can exploit the fact that the product of the diagonal gives me the determinant if the matrix is upper, lower or both triangular- but i don't know how i can turn it into a triangular?
If i try regular row operations, i get:
X(s) =
[s, 1, 1, 1],
[0, s-1, 1, 1],
[0, 0, s-1, 1],
[1-s, 0, 0, s-1]
This is by first subtracting R1 from R2-4, and then subtracting R4 from R2-3. But i can't get any further, how do i solve this?
linear-algebra determinant
asked Jan 30 at 19:50
melbilmelbil
31
$endgroup$
marked as duplicate by Dietrich Burde, Misha Lavrov, GNUSupporter 8964民主女神 地下教會, Cesareo, YiFan Feb 10 at 1:20
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head
$endgroup$
– melbil
Jan 30 at 19:53
$begingroup$
Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting.
$endgroup$
– Dietrich Burde
Jan 30 at 19:54
$begingroup$
I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(mathrm{D}5)$ and $(mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case
$endgroup$
– user635162
Jan 30 at 19:55
$begingroup$
@melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$).
$endgroup$
– Jean Marie
Jan 30 at 20:06
add a comment |
$begingroup$
This question already has an answer here:
Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)
9 answers
Determinant of a matrix will all off-diagonal elements equal
1 answer
i have the following homework problem:
Find the determinant of X(s):
X(s) =
[s, 1, 1, 1],
[1, s, 1, 1],
[1, 1, s, 1],
[1, 1, 1, s]
I know i can exploit the fact that the product of the diagonal gives me the determinant if the matrix is upper, lower or both triangular- but i don't know how i can turn it into a triangular?
If i try regular row operations, i get:
X(s) =
[s, 1, 1, 1],
[0, s-1, 1, 1],
[0, 0, s-1, 1],
[1-s, 0, 0, s-1]
This is by first subtracting R1 from R2-4, and then subtracting R4 from R2-3. But i can't get any further, how do i solve this?
linear-algebra determinant
asked Jan 30 at 19:50
melbilmelbil
31
$endgroup$
This question already has an answer here:
Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)
9 answers
Determinant of a matrix will all off-diagonal elements equal
1 answer
i have the following homework problem:
Find the determinant of X(s):
X(s) =
[s, 1, 1, 1],
[1, s, 1, 1],
[1, 1, s, 1],
[1, 1, 1, s]
I know i can exploit the fact that the product of the diagonal gives me the determinant if the matrix is upper, lower or both triangular- but i don't know how i can turn it into a triangular?
If i try regular row operations, i get:
X(s) =
[s, 1, 1, 1],
[0, s-1, 1, 1],
[0, 0, s-1, 1],
[1-s, 0, 0, s-1]
This is by first subtracting R1 from R2-4, and then subtracting R4 from R2-3. But i can't get any further, how do i solve this?
This question already has an answer here:
Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)
9 answers
Determinant of a matrix will all off-diagonal elements equal
1 answer
linear-algebra determinant
linear-algebra determinant
asked Jan 30 at 19:50
melbilmelbil
31
asked Jan 30 at 19:50
melbilmelbil
31
asked Jan 30 at 19:50
melbilmelbil
31
asked Jan 30 at 19:50
melbilmelbil
31
asked Jan 30 at 19:50
melbilmelbil
31
31
marked as duplicate by Dietrich Burde, Misha Lavrov, GNUSupporter 8964民主女神 地下教會, Cesareo, YiFan Feb 10 at 1:20
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, Misha Lavrov, GNUSupporter 8964民主女神 地下教會, Cesareo, YiFan Feb 10 at 1:20
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head
$endgroup$
– melbil
Jan 30 at 19:53
$begingroup$
Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting.
$endgroup$
– Dietrich Burde
Jan 30 at 19:54
$begingroup$
I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(mathrm{D}5)$ and $(mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case
$endgroup$
– user635162
Jan 30 at 19:55
$begingroup$
@melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$).
$endgroup$
– Jean Marie
Jan 30 at 20:06
add a comment |
$begingroup$
What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head
$endgroup$
– melbil
Jan 30 at 19:53
$begingroup$
Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting.
$endgroup$
– Dietrich Burde
Jan 30 at 19:54
$begingroup$
I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(mathrm{D}5)$ and $(mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case
$endgroup$
– user635162
Jan 30 at 19:55
$begingroup$
@melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$).
$endgroup$
– Jean Marie
Jan 30 at 20:06
$begingroup$
What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head
$endgroup$
– melbil
Jan 30 at 19:53
$begingroup$
What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head
$endgroup$
– melbil
Jan 30 at 19:53
$begingroup$
Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting.
$endgroup$
– Dietrich Burde
Jan 30 at 19:54
$begingroup$
Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting.
$endgroup$
– Dietrich Burde
Jan 30 at 19:54
$begingroup$
I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(mathrm{D}5)$ and $(mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case
$endgroup$
– user635162
Jan 30 at 19:55
$begingroup$
I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(mathrm{D}5)$ and $(mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case
$endgroup$
– user635162
Jan 30 at 19:55
$begingroup$
@melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$).
$endgroup$
– Jean Marie
Jan 30 at 20:06
$begingroup$
@melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$).
$endgroup$
– Jean Marie
Jan 30 at 20:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's something simple: first exploit linearity of the determinant w.r.t. rows:
$$ D=begin{vmatrix}
s&1&1&1\1&s&1&1\1&1&s&1\1&1&1&s
end{vmatrix}=
begin{vmatrix}
s-1&0&0&1-s\0&s-1&0&1-s\0&0&s-1&1-s\1&1&1&s
end{vmatrix}= (s-1)^3
begin{vmatrix}
1&0&0&-1\0&1&0&-1\0&0&1&-1\1&1&1&s
end{vmatrix}$$
Next expand by the first row:
$$D=(s-1)^3left(1cdotbegin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}-(-1)
begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix},right)$$
Now the two $3times3$ determinants are:
begin{align}
&begin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}=begin{vmatrix}
1&0&0\ 0&1&0\1 &1&s+2
end{vmatrix}=s+2\[1ex]
&begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix}=1qquadtext{(expanding by the first column)}
end{align}
and ultimately, we obtain
$$D=(s-1)^3(s+3).$$
answered Jan 30 at 20:34
BernardBernard
124k741116
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$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's something simple: first exploit linearity of the determinant w.r.t. rows:
$$ D=begin{vmatrix}
s&1&1&1\1&s&1&1\1&1&s&1\1&1&1&s
end{vmatrix}=
begin{vmatrix}
s-1&0&0&1-s\0&s-1&0&1-s\0&0&s-1&1-s\1&1&1&s
end{vmatrix}= (s-1)^3
begin{vmatrix}
1&0&0&-1\0&1&0&-1\0&0&1&-1\1&1&1&s
end{vmatrix}$$
Next expand by the first row:
$$D=(s-1)^3left(1cdotbegin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}-(-1)
begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix},right)$$
Now the two $3times3$ determinants are:
begin{align}
&begin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}=begin{vmatrix}
1&0&0\ 0&1&0\1 &1&s+2
end{vmatrix}=s+2\[1ex]
&begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix}=1qquadtext{(expanding by the first column)}
end{align}
and ultimately, we obtain
$$D=(s-1)^3(s+3).$$
answered Jan 30 at 20:34
BernardBernard
124k741116
$endgroup$
$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
add a comment |
$begingroup$
Here's something simple: first exploit linearity of the determinant w.r.t. rows:
$$ D=begin{vmatrix}
s&1&1&1\1&s&1&1\1&1&s&1\1&1&1&s
end{vmatrix}=
begin{vmatrix}
s-1&0&0&1-s\0&s-1&0&1-s\0&0&s-1&1-s\1&1&1&s
end{vmatrix}= (s-1)^3
begin{vmatrix}
1&0&0&-1\0&1&0&-1\0&0&1&-1\1&1&1&s
end{vmatrix}$$
Next expand by the first row:
$$D=(s-1)^3left(1cdotbegin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}-(-1)
begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix},right)$$
Now the two $3times3$ determinants are:
begin{align}
&begin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}=begin{vmatrix}
1&0&0\ 0&1&0\1 &1&s+2
end{vmatrix}=s+2\[1ex]
&begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix}=1qquadtext{(expanding by the first column)}
end{align}
and ultimately, we obtain
$$D=(s-1)^3(s+3).$$
answered Jan 30 at 20:34
BernardBernard
124k741116
$endgroup$
$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
add a comment |
$begingroup$
Here's something simple: first exploit linearity of the determinant w.r.t. rows:
$$ D=begin{vmatrix}
s&1&1&1\1&s&1&1\1&1&s&1\1&1&1&s
end{vmatrix}=
begin{vmatrix}
s-1&0&0&1-s\0&s-1&0&1-s\0&0&s-1&1-s\1&1&1&s
end{vmatrix}= (s-1)^3
begin{vmatrix}
1&0&0&-1\0&1&0&-1\0&0&1&-1\1&1&1&s
end{vmatrix}$$
Next expand by the first row:
$$D=(s-1)^3left(1cdotbegin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}-(-1)
begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix},right)$$
Now the two $3times3$ determinants are:
begin{align}
&begin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}=begin{vmatrix}
1&0&0\ 0&1&0\1 &1&s+2
end{vmatrix}=s+2\[1ex]
&begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix}=1qquadtext{(expanding by the first column)}
end{align}
and ultimately, we obtain
$$D=(s-1)^3(s+3).$$
answered Jan 30 at 20:34
BernardBernard
124k741116
$endgroup$
Here's something simple: first exploit linearity of the determinant w.r.t. rows:
$$ D=begin{vmatrix}
s&1&1&1\1&s&1&1\1&1&s&1\1&1&1&s
end{vmatrix}=
begin{vmatrix}
s-1&0&0&1-s\0&s-1&0&1-s\0&0&s-1&1-s\1&1&1&s
end{vmatrix}= (s-1)^3
begin{vmatrix}
1&0&0&-1\0&1&0&-1\0&0&1&-1\1&1&1&s
end{vmatrix}$$
Next expand by the first row:
$$D=(s-1)^3left(1cdotbegin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}-(-1)
begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix},right)$$
Now the two $3times3$ determinants are:
begin{align}
&begin{vmatrix}
1&0&-1\ 0&1&-1\1 &1&s
end{vmatrix}=begin{vmatrix}
1&0&0\ 0&1&0\1 &1&s+2
end{vmatrix}=s+2\[1ex]
&begin{vmatrix}
0&1&0\0&0&1\1&1&1
end{vmatrix}=1qquadtext{(expanding by the first column)}
end{align}
and ultimately, we obtain
$$D=(s-1)^3(s+3).$$
answered Jan 30 at 20:34
BernardBernard
124k741116
answered Jan 30 at 20:34
BernardBernard
124k741116
answered Jan 30 at 20:34
BernardBernard
124k741116
answered Jan 30 at 20:34
BernardBernard
124k741116
124k741116
$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
add a comment |
$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
Thank you very much for taking the time to explain that.
$endgroup$
– melbil
Jan 30 at 21:37
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
$begingroup$
You still have to find out what linear combinations were done ;o)
$endgroup$
– Bernard
Jan 30 at 21:39
add a comment |
$begingroup$
What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head
$endgroup$
– melbil
Jan 30 at 19:53
$begingroup$
Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting.
$endgroup$
– Dietrich Burde
Jan 30 at 19:54
$begingroup$
I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(mathrm{D}5)$ and $(mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case
$endgroup$
– user635162
Jan 30 at 19:55
$begingroup$
@melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$).
$endgroup$
– Jean Marie
Jan 30 at 20:06