4x4 Determinant [duplicate]
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What will be the value of the following determinant without expanding it?
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$A =beginvmatrixa^2 & b^2 & c^2 & d^2\ (a-1)^2 & (b-1)^2 & (c-1)^2 & (d-1)^2 \ (a-2)^2 & (b-2)^2 & (c-2)^2 & (d-2)^2\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
I have been trying to solve this determinant for a while now and haven't gotten very far.
I have been using row reduction to try and create a lower triangle but keep getting caught up with the amount of ugly multiplication involved and cant seem to find an easier way.
Any help with this question would be much appreciated.
linear-algebra determinant
edited Dec 10 at 8:03
Jean Marie
28.7k41849
asked Dec 9 at 21:24
Batmobile
715
marked as duplicate by Martin Sleziak, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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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.
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up vote
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down vote
favorite
This question already has an answer here:
What will be the value of the following determinant without expanding it?
7 answers
$A =beginvmatrixa^2 & b^2 & c^2 & d^2\ (a-1)^2 & (b-1)^2 & (c-1)^2 & (d-1)^2 \ (a-2)^2 & (b-2)^2 & (c-2)^2 & (d-2)^2\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
I have been trying to solve this determinant for a while now and haven't gotten very far.
I have been using row reduction to try and create a lower triangle but keep getting caught up with the amount of ugly multiplication involved and cant seem to find an easier way.
Any help with this question would be much appreciated.
linear-algebra determinant
edited Dec 10 at 8:03
Jean Marie
28.7k41849
asked Dec 9 at 21:24
Batmobile
715
marked as duplicate by Martin Sleziak, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Dec 10 at 17:21
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.
2
To gain intuition on problems like these, you could always plug in values for $a, b, c, d$ and see what happens. You would have quickly come to the conjecture that the determinant is always zero, after which you could have concentrated on proving that.
– John Coleman
Dec 10 at 12:39
Comment here mentions that this determinant appears as an exercise in Golan's book (and probably in many other places): What will be the value of the following determinant without expanding it?
– Martin Sleziak
Dec 10 at 17:06
add a comment |
up vote
11
down vote
favorite
up vote
11
down vote
favorite
This question already has an answer here:
What will be the value of the following determinant without expanding it?
7 answers
$A =beginvmatrixa^2 & b^2 & c^2 & d^2\ (a-1)^2 & (b-1)^2 & (c-1)^2 & (d-1)^2 \ (a-2)^2 & (b-2)^2 & (c-2)^2 & (d-2)^2\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
I have been trying to solve this determinant for a while now and haven't gotten very far.
I have been using row reduction to try and create a lower triangle but keep getting caught up with the amount of ugly multiplication involved and cant seem to find an easier way.
Any help with this question would be much appreciated.
linear-algebra determinant
edited Dec 10 at 8:03
Jean Marie
28.7k41849
asked Dec 9 at 21:24
Batmobile
715
This question already has an answer here:
What will be the value of the following determinant without expanding it?
7 answers
$A =beginvmatrixa^2 & b^2 & c^2 & d^2\ (a-1)^2 & (b-1)^2 & (c-1)^2 & (d-1)^2 \ (a-2)^2 & (b-2)^2 & (c-2)^2 & (d-2)^2\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
I have been trying to solve this determinant for a while now and haven't gotten very far.
I have been using row reduction to try and create a lower triangle but keep getting caught up with the amount of ugly multiplication involved and cant seem to find an easier way.
Any help with this question would be much appreciated.
This question already has an answer here:
What will be the value of the following determinant without expanding it?
7 answers
linear-algebra determinant
linear-algebra determinant
edited Dec 10 at 8:03
Jean Marie
28.7k41849
asked Dec 9 at 21:24
Batmobile
715
edited Dec 10 at 8:03
Jean Marie
28.7k41849
asked Dec 9 at 21:24
Batmobile
715
edited Dec 10 at 8:03
Jean Marie
28.7k41849
edited Dec 10 at 8:03
Jean Marie
28.7k41849
edited Dec 10 at 8:03
Jean Marie
28.7k41849
28.7k41849
asked Dec 9 at 21:24
Batmobile
715
asked Dec 9 at 21:24
Batmobile
715
asked Dec 9 at 21:24
Batmobile
715
715
marked as duplicate by Martin Sleziak, José Carlos Santos linear-algebra
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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.
2
To gain intuition on problems like these, you could always plug in values for $a, b, c, d$ and see what happens. You would have quickly come to the conjecture that the determinant is always zero, after which you could have concentrated on proving that.
– John Coleman
Dec 10 at 12:39
Comment here mentions that this determinant appears as an exercise in Golan's book (and probably in many other places): What will be the value of the following determinant without expanding it?
– Martin Sleziak
Dec 10 at 17:06
add a comment |
2
To gain intuition on problems like these, you could always plug in values for $a, b, c, d$ and see what happens. You would have quickly come to the conjecture that the determinant is always zero, after which you could have concentrated on proving that.
– John Coleman
Dec 10 at 12:39
Comment here mentions that this determinant appears as an exercise in Golan's book (and probably in many other places): What will be the value of the following determinant without expanding it?
– Martin Sleziak
Dec 10 at 17:06
2
2
To gain intuition on problems like these, you could always plug in values for $a, b, c, d$ and see what happens. You would have quickly come to the conjecture that the determinant is always zero, after which you could have concentrated on proving that.
– John Coleman
Dec 10 at 12:39
To gain intuition on problems like these, you could always plug in values for $a, b, c, d$ and see what happens. You would have quickly come to the conjecture that the determinant is always zero, after which you could have concentrated on proving that.
– John Coleman
Dec 10 at 12:39
Comment here mentions that this determinant appears as an exercise in Golan's book (and probably in many other places): What will be the value of the following determinant without expanding it?
– Martin Sleziak
Dec 10 at 17:06
Comment here mentions that this determinant appears as an exercise in Golan's book (and probably in many other places): What will be the value of the following determinant without expanding it?
– Martin Sleziak
Dec 10 at 17:06
add a comment |
2 Answers
2
active
oldest
votes
up vote
21
down vote
accepted
Use row transformations in the following manner:
$R_1to R_1-R_2\R_2to R_2-R_3\R_3to R_3-R_4$
$=beginvmatrix2a-1 & 2b-1 & 2c-1 & 2d-1\ 2a-3 & 2b-3 & 2c-3 & 2d-3 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
$R_1to R_1-R_2\R_2to R_2-R_3$
$=beginvmatrix2 & 2 & 2 & 2\ 2 & 2 & 2 & 2 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
Determinant having identical rows evaluates to $0$. You could go an extra mile and perform $R_1to R_1-R_2$ to see why this is true.
answered Dec 9 at 21:32
Shubham Johri
3,019413
add a comment |
up vote
23
down vote
@Shubham Johri : There is a simple proof that this determinant, or more precisely its transpose, is zero.
Let us consider the following matrix-vector product :
$$beginbmatrix
a^2 & (a-1)^2 & (a-2)^2 & (a-3)^2\
b^2 & (b-1)^2 & (b-2)^2 & (b-3)^2\
c^2 & (c-1)^2 & (c-2)^2 & (c-3)^2\
d^2 & (d-1)^2 & (d-2)^2 & (d-3)^2
endbmatrix
beginbmatrix
-1\ 3\-3\ 1
endbmatrix=beginbmatrix
0\0\0\0
endbmatrix.$$
Thus this matrix has a non-zero kernel ; therefore its determinant is equal to $0$.
Remark : it wasn't necessary to explicitate the vector with coordinates $-1,3,-3,1$ (have you noticed that it uses binomial coefficients ?) ; we could simply have said that such a vector with $p,q,r,s$ coordinates exist because in the 3-dimensional vector space $mathbbR[X]_deg leq 2$, any set of more than 3 vectors is necessarily linked by a zero linear combination :
$$p X^2 + q (X-1)^2 + r (X-2)^2 + s (X-3)^2 = 0.$$
answered Dec 9 at 22:54
Jean Marie
28.7k41849
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
21
down vote
accepted
Use row transformations in the following manner:
$R_1to R_1-R_2\R_2to R_2-R_3\R_3to R_3-R_4$
$=beginvmatrix2a-1 & 2b-1 & 2c-1 & 2d-1\ 2a-3 & 2b-3 & 2c-3 & 2d-3 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
$R_1to R_1-R_2\R_2to R_2-R_3$
$=beginvmatrix2 & 2 & 2 & 2\ 2 & 2 & 2 & 2 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
Determinant having identical rows evaluates to $0$. You could go an extra mile and perform $R_1to R_1-R_2$ to see why this is true.
answered Dec 9 at 21:32
Shubham Johri
3,019413
add a comment |
up vote
21
down vote
accepted
Use row transformations in the following manner:
$R_1to R_1-R_2\R_2to R_2-R_3\R_3to R_3-R_4$
$=beginvmatrix2a-1 & 2b-1 & 2c-1 & 2d-1\ 2a-3 & 2b-3 & 2c-3 & 2d-3 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
$R_1to R_1-R_2\R_2to R_2-R_3$
$=beginvmatrix2 & 2 & 2 & 2\ 2 & 2 & 2 & 2 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
Determinant having identical rows evaluates to $0$. You could go an extra mile and perform $R_1to R_1-R_2$ to see why this is true.
answered Dec 9 at 21:32
Shubham Johri
3,019413
add a comment |
up vote
21
down vote
accepted
up vote
21
down vote
accepted
Use row transformations in the following manner:
$R_1to R_1-R_2\R_2to R_2-R_3\R_3to R_3-R_4$
$=beginvmatrix2a-1 & 2b-1 & 2c-1 & 2d-1\ 2a-3 & 2b-3 & 2c-3 & 2d-3 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
$R_1to R_1-R_2\R_2to R_2-R_3$
$=beginvmatrix2 & 2 & 2 & 2\ 2 & 2 & 2 & 2 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
Determinant having identical rows evaluates to $0$. You could go an extra mile and perform $R_1to R_1-R_2$ to see why this is true.
answered Dec 9 at 21:32
Shubham Johri
3,019413
Use row transformations in the following manner:
$R_1to R_1-R_2\R_2to R_2-R_3\R_3to R_3-R_4$
$=beginvmatrix2a-1 & 2b-1 & 2c-1 & 2d-1\ 2a-3 & 2b-3 & 2c-3 & 2d-3 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
$R_1to R_1-R_2\R_2to R_2-R_3$
$=beginvmatrix2 & 2 & 2 & 2\ 2 & 2 & 2 & 2 \ 2a-5 & 2b-5 & 2c-5 & 2d-5\ (a-3)^2 & (b-3)^2 & (c-3)^2 & (d-3)^2endvmatrix$
Determinant having identical rows evaluates to $0$. You could go an extra mile and perform $R_1to R_1-R_2$ to see why this is true.
answered Dec 9 at 21:32
Shubham Johri
3,019413
answered Dec 9 at 21:32
Shubham Johri
3,019413
answered Dec 9 at 21:32
Shubham Johri
3,019413
answered Dec 9 at 21:32
Shubham Johri
3,019413
3,019413
add a comment |
add a comment |
up vote
23
down vote
@Shubham Johri : There is a simple proof that this determinant, or more precisely its transpose, is zero.
Let us consider the following matrix-vector product :
$$beginbmatrix
a^2 & (a-1)^2 & (a-2)^2 & (a-3)^2\
b^2 & (b-1)^2 & (b-2)^2 & (b-3)^2\
c^2 & (c-1)^2 & (c-2)^2 & (c-3)^2\
d^2 & (d-1)^2 & (d-2)^2 & (d-3)^2
endbmatrix
beginbmatrix
-1\ 3\-3\ 1
endbmatrix=beginbmatrix
0\0\0\0
endbmatrix.$$
Thus this matrix has a non-zero kernel ; therefore its determinant is equal to $0$.
Remark : it wasn't necessary to explicitate the vector with coordinates $-1,3,-3,1$ (have you noticed that it uses binomial coefficients ?) ; we could simply have said that such a vector with $p,q,r,s$ coordinates exist because in the 3-dimensional vector space $mathbbR[X]_deg leq 2$, any set of more than 3 vectors is necessarily linked by a zero linear combination :
$$p X^2 + q (X-1)^2 + r (X-2)^2 + s (X-3)^2 = 0.$$
answered Dec 9 at 22:54
Jean Marie
28.7k41849
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
add a comment |
up vote
23
down vote
@Shubham Johri : There is a simple proof that this determinant, or more precisely its transpose, is zero.
Let us consider the following matrix-vector product :
$$beginbmatrix
a^2 & (a-1)^2 & (a-2)^2 & (a-3)^2\
b^2 & (b-1)^2 & (b-2)^2 & (b-3)^2\
c^2 & (c-1)^2 & (c-2)^2 & (c-3)^2\
d^2 & (d-1)^2 & (d-2)^2 & (d-3)^2
endbmatrix
beginbmatrix
-1\ 3\-3\ 1
endbmatrix=beginbmatrix
0\0\0\0
endbmatrix.$$
Thus this matrix has a non-zero kernel ; therefore its determinant is equal to $0$.
Remark : it wasn't necessary to explicitate the vector with coordinates $-1,3,-3,1$ (have you noticed that it uses binomial coefficients ?) ; we could simply have said that such a vector with $p,q,r,s$ coordinates exist because in the 3-dimensional vector space $mathbbR[X]_deg leq 2$, any set of more than 3 vectors is necessarily linked by a zero linear combination :
$$p X^2 + q (X-1)^2 + r (X-2)^2 + s (X-3)^2 = 0.$$
answered Dec 9 at 22:54
Jean Marie
28.7k41849
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
add a comment |
up vote
23
down vote
up vote
23
down vote
@Shubham Johri : There is a simple proof that this determinant, or more precisely its transpose, is zero.
Let us consider the following matrix-vector product :
$$beginbmatrix
a^2 & (a-1)^2 & (a-2)^2 & (a-3)^2\
b^2 & (b-1)^2 & (b-2)^2 & (b-3)^2\
c^2 & (c-1)^2 & (c-2)^2 & (c-3)^2\
d^2 & (d-1)^2 & (d-2)^2 & (d-3)^2
endbmatrix
beginbmatrix
-1\ 3\-3\ 1
endbmatrix=beginbmatrix
0\0\0\0
endbmatrix.$$
Thus this matrix has a non-zero kernel ; therefore its determinant is equal to $0$.
Remark : it wasn't necessary to explicitate the vector with coordinates $-1,3,-3,1$ (have you noticed that it uses binomial coefficients ?) ; we could simply have said that such a vector with $p,q,r,s$ coordinates exist because in the 3-dimensional vector space $mathbbR[X]_deg leq 2$, any set of more than 3 vectors is necessarily linked by a zero linear combination :
$$p X^2 + q (X-1)^2 + r (X-2)^2 + s (X-3)^2 = 0.$$
answered Dec 9 at 22:54
Jean Marie
28.7k41849
@Shubham Johri : There is a simple proof that this determinant, or more precisely its transpose, is zero.
Let us consider the following matrix-vector product :
$$beginbmatrix
a^2 & (a-1)^2 & (a-2)^2 & (a-3)^2\
b^2 & (b-1)^2 & (b-2)^2 & (b-3)^2\
c^2 & (c-1)^2 & (c-2)^2 & (c-3)^2\
d^2 & (d-1)^2 & (d-2)^2 & (d-3)^2
endbmatrix
beginbmatrix
-1\ 3\-3\ 1
endbmatrix=beginbmatrix
0\0\0\0
endbmatrix.$$
Thus this matrix has a non-zero kernel ; therefore its determinant is equal to $0$.
Remark : it wasn't necessary to explicitate the vector with coordinates $-1,3,-3,1$ (have you noticed that it uses binomial coefficients ?) ; we could simply have said that such a vector with $p,q,r,s$ coordinates exist because in the 3-dimensional vector space $mathbbR[X]_deg leq 2$, any set of more than 3 vectors is necessarily linked by a zero linear combination :
$$p X^2 + q (X-1)^2 + r (X-2)^2 + s (X-3)^2 = 0.$$
answered Dec 9 at 22:54
Jean Marie
28.7k41849
edited Dec 10 at 5:40
answered Dec 9 at 22:54
Jean Marie
28.7k41849
answered Dec 9 at 22:54
Jean Marie
28.7k41849
answered Dec 9 at 22:54
Jean Marie
28.7k41849
28.7k41849
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
add a comment |
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
That's a nice way to solve the question too
– Shubham Johri
Dec 10 at 8:48
add a comment |
2
To gain intuition on problems like these, you could always plug in values for $a, b, c, d$ and see what happens. You would have quickly come to the conjecture that the determinant is always zero, after which you could have concentrated on proving that.
– John Coleman
Dec 10 at 12:39
Comment here mentions that this determinant appears as an exercise in Golan's book (and probably in many other places): What will be the value of the following determinant without expanding it?
– Martin Sleziak
Dec 10 at 17:06