Mixing
Determinant of matrix that contains a column of same numbers
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Given $4times 4$ matrix:
$$beginvmatrix
x & x+a& 0& 1\
x+a& x& a& 1\
0& a& x& 1\
1& -1& 1& 1
endvmatrix=x+a$$
Is it safe to assume that the determinant equals $0$ because there is a column than contains the same numbers(most right) and using Gaussian elimination technique I can turn it into zeroes and then the determinant is $0$. Is it correct?
linear-algebra matrices determinant
edited Jul 15 at 19:46
copper.hat
122k557156
asked Jul 15 at 19:05
user6394019
30311
 |Â
show 3 more comments
up vote
0
down vote
favorite
Given $4times 4$ matrix:
$$beginvmatrix
x & x+a& 0& 1\
x+a& x& a& 1\
0& a& x& 1\
1& -1& 1& 1
endvmatrix=x+a$$
Is it safe to assume that the determinant equals $0$ because there is a column than contains the same numbers(most right) and using Gaussian elimination technique I can turn it into zeroes and then the determinant is $0$. Is it correct?
linear-algebra matrices determinant
edited Jul 15 at 19:46
copper.hat
122k557156
asked Jul 15 at 19:05
user6394019
30311
6
What is the determinant of $beginbmatrix 1 & a \ 1 & b endbmatrix$?
– Nitin
Jul 15 at 19:07
Depends on the number. A column of zero elements would zero the determinant. Otherwise the four columns vectors might (but must not) be linear independent, which would give a non-zero determinant.
– mvw
Jul 15 at 19:12
The determinant is neither zero nor $x+a$. You should fix or reword the question.
– copper.hat
Jul 15 at 19:47
@copper.hat sorry to dissapoint you this is the determinant. Our teacher solved it today in class
– user6394019
Jul 15 at 20:11
You might want to check again that your question is the one that your teacher solved.
– copper.hat
Jul 15 at 20:19
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $4times 4$ matrix:
$$beginvmatrix
x & x+a& 0& 1\
x+a& x& a& 1\
0& a& x& 1\
1& -1& 1& 1
endvmatrix=x+a$$
Is it safe to assume that the determinant equals $0$ because there is a column than contains the same numbers(most right) and using Gaussian elimination technique I can turn it into zeroes and then the determinant is $0$. Is it correct?
linear-algebra matrices determinant
edited Jul 15 at 19:46
copper.hat
122k557156
asked Jul 15 at 19:05
user6394019
30311
Given $4times 4$ matrix:
$$beginvmatrix
x & x+a& 0& 1\
x+a& x& a& 1\
0& a& x& 1\
1& -1& 1& 1
endvmatrix=x+a$$
Is it safe to assume that the determinant equals $0$ because there is a column than contains the same numbers(most right) and using Gaussian elimination technique I can turn it into zeroes and then the determinant is $0$. Is it correct?
linear-algebra matrices determinant
edited Jul 15 at 19:46
copper.hat
122k557156
asked Jul 15 at 19:05
user6394019
30311
edited Jul 15 at 19:46
copper.hat
122k557156
edited Jul 15 at 19:46
copper.hat
122k557156
edited Jul 15 at 19:46
copper.hat
122k557156
122k557156
asked Jul 15 at 19:05
user6394019
30311
asked Jul 15 at 19:05
user6394019
30311
asked Jul 15 at 19:05
user6394019
30311
30311
6
What is the determinant of $beginbmatrix 1 & a \ 1 & b endbmatrix$?
– Nitin
Jul 15 at 19:07
Depends on the number. A column of zero elements would zero the determinant. Otherwise the four columns vectors might (but must not) be linear independent, which would give a non-zero determinant.
– mvw
Jul 15 at 19:12
The determinant is neither zero nor $x+a$. You should fix or reword the question.
– copper.hat
Jul 15 at 19:47
@copper.hat sorry to dissapoint you this is the determinant. Our teacher solved it today in class
– user6394019
Jul 15 at 20:11
You might want to check again that your question is the one that your teacher solved.
– copper.hat
Jul 15 at 20:19
 |Â
show 3 more comments
6
What is the determinant of $beginbmatrix 1 & a \ 1 & b endbmatrix$?
– Nitin
Jul 15 at 19:07
Depends on the number. A column of zero elements would zero the determinant. Otherwise the four columns vectors might (but must not) be linear independent, which would give a non-zero determinant.
– mvw
Jul 15 at 19:12
The determinant is neither zero nor $x+a$. You should fix or reword the question.
– copper.hat
Jul 15 at 19:47
@copper.hat sorry to dissapoint you this is the determinant. Our teacher solved it today in class
– user6394019
Jul 15 at 20:11
You might want to check again that your question is the one that your teacher solved.
– copper.hat
Jul 15 at 20:19
6
6
What is the determinant of $beginbmatrix 1 & a \ 1 & b endbmatrix$?
– Nitin
Jul 15 at 19:07
What is the determinant of $beginbmatrix 1 & a \ 1 & b endbmatrix$?
– Nitin
Jul 15 at 19:07
Depends on the number. A column of zero elements would zero the determinant. Otherwise the four columns vectors might (but must not) be linear independent, which would give a non-zero determinant.
– mvw
Jul 15 at 19:12
Depends on the number. A column of zero elements would zero the determinant. Otherwise the four columns vectors might (but must not) be linear independent, which would give a non-zero determinant.
– mvw
Jul 15 at 19:12
The determinant is neither zero nor $x+a$. You should fix or reword the question.
– copper.hat
Jul 15 at 19:47
The determinant is neither zero nor $x+a$. You should fix or reword the question.
– copper.hat
Jul 15 at 19:47
@copper.hat sorry to dissapoint you this is the determinant. Our teacher solved it today in class
– user6394019
Jul 15 at 20:11
@copper.hat sorry to dissapoint you this is the determinant. Our teacher solved it today in class
– user6394019
Jul 15 at 20:11
You might want to check again that your question is the one that your teacher solved.
– copper.hat
Jul 15 at 20:19
You might want to check again that your question is the one that your teacher solved.
– copper.hat
Jul 15 at 20:19
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
No. Try to think about what happens if you were to carry out the Gaussian elimination you propose. You can turn all but one of the $1$s into a $0$, but your proposed last step wouldn't work -- you can't subtract a row from itself when doing Gaussian elimination.
answered Jul 15 at 19:59
JonathanZ
1,702412
add a comment |Â
up vote
0
down vote
No, it is not correct, unless that “same number†is $0$. For instance$$beginvmatrix1&0&0&ldots&0&a\0&1&0&ldots&0&a\0&0&1&ldots&0&a\vdots&vdots&vdots&ddots&vdots&vdots\0&0&0&cdots&1&a\0&0&0&cdots&0&aendvmatrix=a.$$
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
Fix $a$ and let $p$ be the value of the determinant which is a polynomial in $x$. Note that $p(0) = p(-a) = 0$ and so $x(x+a)$ must divide $p$. Also note that $p$ has degree 2.
A little more grind shows that $p(a)=-2(1+a)^2$ from which we conclude that $p(x) = -2(1+a)x(x+a)$.
Alternatively, let WA do the work.
answered Jul 15 at 19:27
copper.hat
122k557156
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
No. Try to think about what happens if you were to carry out the Gaussian elimination you propose. You can turn all but one of the $1$s into a $0$, but your proposed last step wouldn't work -- you can't subtract a row from itself when doing Gaussian elimination.
answered Jul 15 at 19:59
JonathanZ
1,702412
add a comment |Â
up vote
1
down vote
No. Try to think about what happens if you were to carry out the Gaussian elimination you propose. You can turn all but one of the $1$s into a $0$, but your proposed last step wouldn't work -- you can't subtract a row from itself when doing Gaussian elimination.
answered Jul 15 at 19:59
JonathanZ
1,702412
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No. Try to think about what happens if you were to carry out the Gaussian elimination you propose. You can turn all but one of the $1$s into a $0$, but your proposed last step wouldn't work -- you can't subtract a row from itself when doing Gaussian elimination.
answered Jul 15 at 19:59
JonathanZ
1,702412
No. Try to think about what happens if you were to carry out the Gaussian elimination you propose. You can turn all but one of the $1$s into a $0$, but your proposed last step wouldn't work -- you can't subtract a row from itself when doing Gaussian elimination.
answered Jul 15 at 19:59
JonathanZ
1,702412
answered Jul 15 at 19:59
JonathanZ
1,702412
answered Jul 15 at 19:59
JonathanZ
1,702412
answered Jul 15 at 19:59
JonathanZ
1,702412
1,702412
add a comment |Â
add a comment |Â
up vote
0
down vote
No, it is not correct, unless that “same number†is $0$. For instance$$beginvmatrix1&0&0&ldots&0&a\0&1&0&ldots&0&a\0&0&1&ldots&0&a\vdots&vdots&vdots&ddots&vdots&vdots\0&0&0&cdots&1&a\0&0&0&cdots&0&aendvmatrix=a.$$
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
No, it is not correct, unless that “same number†is $0$. For instance$$beginvmatrix1&0&0&ldots&0&a\0&1&0&ldots&0&a\0&0&1&ldots&0&a\vdots&vdots&vdots&ddots&vdots&vdots\0&0&0&cdots&1&a\0&0&0&cdots&0&aendvmatrix=a.$$
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
add a comment |Â
up vote
0
down vote
up vote
0
down vote
No, it is not correct, unless that “same number†is $0$. For instance$$beginvmatrix1&0&0&ldots&0&a\0&1&0&ldots&0&a\0&0&1&ldots&0&a\vdots&vdots&vdots&ddots&vdots&vdots\0&0&0&cdots&1&a\0&0&0&cdots&0&aendvmatrix=a.$$
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
No, it is not correct, unless that “same number†is $0$. For instance$$beginvmatrix1&0&0&ldots&0&a\0&1&0&ldots&0&a\0&0&1&ldots&0&a\vdots&vdots&vdots&ddots&vdots&vdots\0&0&0&cdots&1&a\0&0&0&cdots&0&aendvmatrix=a.$$
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
answered Jul 15 at 19:26
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
0
down vote
Fix $a$ and let $p$ be the value of the determinant which is a polynomial in $x$. Note that $p(0) = p(-a) = 0$ and so $x(x+a)$ must divide $p$. Also note that $p$ has degree 2.
A little more grind shows that $p(a)=-2(1+a)^2$ from which we conclude that $p(x) = -2(1+a)x(x+a)$.
Alternatively, let WA do the work.
answered Jul 15 at 19:27
copper.hat
122k557156
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
add a comment |Â
up vote
0
down vote
Fix $a$ and let $p$ be the value of the determinant which is a polynomial in $x$. Note that $p(0) = p(-a) = 0$ and so $x(x+a)$ must divide $p$. Also note that $p$ has degree 2.
A little more grind shows that $p(a)=-2(1+a)^2$ from which we conclude that $p(x) = -2(1+a)x(x+a)$.
Alternatively, let WA do the work.
answered Jul 15 at 19:27
copper.hat
122k557156
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Fix $a$ and let $p$ be the value of the determinant which is a polynomial in $x$. Note that $p(0) = p(-a) = 0$ and so $x(x+a)$ must divide $p$. Also note that $p$ has degree 2.
A little more grind shows that $p(a)=-2(1+a)^2$ from which we conclude that $p(x) = -2(1+a)x(x+a)$.
Alternatively, let WA do the work.
answered Jul 15 at 19:27
copper.hat
122k557156
Fix $a$ and let $p$ be the value of the determinant which is a polynomial in $x$. Note that $p(0) = p(-a) = 0$ and so $x(x+a)$ must divide $p$. Also note that $p$ has degree 2.
A little more grind shows that $p(a)=-2(1+a)^2$ from which we conclude that $p(x) = -2(1+a)x(x+a)$.
Alternatively, let WA do the work.
answered Jul 15 at 19:27
copper.hat
122k557156
edited Jul 16 at 2:41
answered Jul 15 at 19:27
copper.hat
122k557156
answered Jul 15 at 19:27
copper.hat
122k557156
answered Jul 15 at 19:27
copper.hat
122k557156
122k557156
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
add a comment |Â
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
I can't quickly see that $p$ has degrees 2; there's an $x^3$ term along the main diagonal if you evaluate it simple-mindedly. Can you add some more details on that point?
– JonathanZ
Jul 16 at 1:05
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
@JonathanZ: Originally I thought the same. The only way to get a power of 3 is for the $det$ expansion to go through the $x$ at the $3,3$ element of matrix. However, the $x^2$ terms from the $(1-2),(1-2)$ part of the matrix cancel, leaving only $x^2$ terms in the final expansion.
– copper.hat
Jul 16 at 2:32
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
Okay, I see it now, but I don't think it's obvious and could use a little more detail (like your comment provides).
– JonathanZ
Jul 16 at 3:02
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852772%2fdeterminant-of-matrix-that-contains-a-column-of-same-numbers%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
6
What is the determinant of $beginbmatrix 1 & a \ 1 & b endbmatrix$?
– Nitin
Jul 15 at 19:07
Depends on the number. A column of zero elements would zero the determinant. Otherwise the four columns vectors might (but must not) be linear independent, which would give a non-zero determinant.
– mvw
Jul 15 at 19:12
The determinant is neither zero nor $x+a$. You should fix or reword the question.
– copper.hat
Jul 15 at 19:47
@copper.hat sorry to dissapoint you this is the determinant. Our teacher solved it today in class
– user6394019
Jul 15 at 20:11
You might want to check again that your question is the one that your teacher solved.
– copper.hat
Jul 15 at 20:19