Determinant of an n x n matrix
I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?
This is the matrix :
$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$
linear-algebra matrices determinant
edited 3 hours ago
StubbornAtom
5,26711138
asked 3 hours ago
Paul Vinur
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?
This is the matrix :
$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$
linear-algebra matrices determinant
edited 3 hours ago
StubbornAtom
5,26711138
asked 3 hours ago
Paul Vinur
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
3 hours ago
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
3 hours ago
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
3 hours ago
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
3 hours ago
I did, the determinants for values of n = 3,4,5.. I can sense the pattern I just cannot find the general rule. It would be easier to try simplifying the matrix first. But what do you mean by expanding over the first column?
– Paul Vinur
1 hour ago
add a comment |
I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?
This is the matrix :
$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$
linear-algebra matrices determinant
edited 3 hours ago
StubbornAtom
5,26711138
asked 3 hours ago
Paul Vinur
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?
This is the matrix :
$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$
linear-algebra matrices determinant
linear-algebra matrices determinant
edited 3 hours ago
StubbornAtom
5,26711138
asked 3 hours ago
Paul Vinur
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
StubbornAtom
5,26711138
asked 3 hours ago
Paul Vinur
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
StubbornAtom
5,26711138
edited 3 hours ago
StubbornAtom
5,26711138
edited 3 hours ago
StubbornAtom
5,26711138
5,26711138
asked 3 hours ago
Paul Vinur
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Paul Vinur
264
asked 3 hours ago
Paul Vinur
264
264
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
3 hours ago
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
3 hours ago
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
3 hours ago
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
3 hours ago
I did, the determinants for values of n = 3,4,5.. I can sense the pattern I just cannot find the general rule. It would be easier to try simplifying the matrix first. But what do you mean by expanding over the first column?
– Paul Vinur
1 hour ago
add a comment |
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
3 hours ago
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
3 hours ago
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
3 hours ago
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
3 hours ago
I did, the determinants for values of n = 3,4,5.. I can sense the pattern I just cannot find the general rule. It would be easier to try simplifying the matrix first. But what do you mean by expanding over the first column?
– Paul Vinur
1 hour ago
1
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
3 hours ago
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
3 hours ago
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
3 hours ago
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
3 hours ago
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
3 hours ago
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
3 hours ago
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
3 hours ago
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
3 hours ago
I did, the determinants for values of n = 3,4,5.. I can sense the pattern I just cannot find the general rule. It would be easier to try simplifying the matrix first. But what do you mean by expanding over the first column?
– Paul Vinur
1 hour ago
I did, the determinants for values of n = 3,4,5.. I can sense the pattern I just cannot find the general rule. It would be easier to try simplifying the matrix first. But what do you mean by expanding over the first column?
– Paul Vinur
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.
Then $det(A) = det(B) + e^TB^{-1}e$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.
This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.
answered 2 hours ago
Hans Engler
10.1k11836
1
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
add a comment |
Let $M_n$ be your matrix.
Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you
Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.
This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the leftSubtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).
This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.
After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$
From this, you can deduce
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
answered 53 mins ago
achille hui
95.3k5129256
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.
Then $det(A) = det(B) + e^TB^{-1}e$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.
This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.
answered 2 hours ago
Hans Engler
10.1k11836
1
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
add a comment |
This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.
Then $det(A) = det(B) + e^TB^{-1}e$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.
This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.
answered 2 hours ago
Hans Engler
10.1k11836
1
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
add a comment |
This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.
Then $det(A) = det(B) + e^TB^{-1}e$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.
This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.
answered 2 hours ago
Hans Engler
10.1k11836
This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.
Then $det(A) = det(B) + e^TB^{-1}e$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.
This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.
answered 2 hours ago
Hans Engler
10.1k11836
answered 2 hours ago
Hans Engler
10.1k11836
answered 2 hours ago
Hans Engler
10.1k11836
answered 2 hours ago
Hans Engler
10.1k11836
10.1k11836
1
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
add a comment |
1
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
1
1
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
Isn't $det A=det B (1+e^TB^{-1}e)$ ?
– StubbornAtom
1 hour ago
add a comment |
Let $M_n$ be your matrix.
Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you
Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.
This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the leftSubtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).
This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.
After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$
From this, you can deduce
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
answered 53 mins ago
achille hui
95.3k5129256
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
add a comment |
Let $M_n$ be your matrix.
Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you
Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.
This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the leftSubtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).
This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.
After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$
From this, you can deduce
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
answered 53 mins ago
achille hui
95.3k5129256
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
add a comment |
Let $M_n$ be your matrix.
Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you
Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.
This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the leftSubtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).
This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.
After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$
From this, you can deduce
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
answered 53 mins ago
achille hui
95.3k5129256
Let $M_n$ be your matrix.
Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you
Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.
This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the leftSubtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).
This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.
After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$
From this, you can deduce
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
answered 53 mins ago
achille hui
95.3k5129256
edited 19 mins ago
answered 53 mins ago
achille hui
95.3k5129256
answered 53 mins ago
achille hui
95.3k5129256
answered 53 mins ago
achille hui
95.3k5129256
95.3k5129256
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
add a comment |
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
6 mins ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_ii$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
1 min ago
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Paul Vinur is a new contributor. Be nice, and check out our Code of Conduct.
Paul Vinur is a new contributor. Be nice, and check out our Code of Conduct.
Paul Vinur is a new contributor. Be nice, and check out our Code of Conduct.
Paul Vinur is a new contributor. Be nice, and check out our Code of Conduct.
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1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
3 hours ago
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
3 hours ago
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
3 hours ago
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
3 hours ago
I did, the determinants for values of n = 3,4,5.. I can sense the pattern I just cannot find the general rule. It would be easier to try simplifying the matrix first. But what do you mean by expanding over the first column?
– Paul Vinur
1 hour ago