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Determinant of a particular large matrix

Linearity of the determinantBlock Matrix Zero Determinant Implication?Breaking a determinant into eight piecesDeterminant of a square matrix with a particular patternDeterminant of block matrix with null row vectorDeterminant of a block matrix $2n$ by $2n$Proofs of Determinants of Block matricesHelp Determinant Binary MatrixDeterminant of block matrix - Finding a fieldIf the determinants of two matrices, $A$ and $B$, of order $n$, $= 0$, are they row-equivalent?

0

1

$begingroup$

I've been trying to solve this problem but I stucked.
Let $A = [a_{ij}]$ with size $ 2011 times 2011$ , and given the condition below

begin{equation}a_{ij}= begin{cases}
(-1)^{|i-j|}, & text{if} i neq j \
2, & text{if} i = j
end{cases}
end{equation}
Find $det(A)$. I was thinking to either form block matrices or do row operation, but it became a mess. Please help me to figure this out.

linear-algebra matrices determinant

share|cite|improve this question

edited Mar 12 at 5:13

Travis

63.3k768150

asked Mar 12 at 4:22

SutMarSutMar

1017

$endgroup$

add a comment | 

0

1

$begingroup$

I've been trying to solve this problem but I stucked.
Let $A = [a_{ij}]$ with size $ 2011 times 2011$ , and given the condition below

begin{equation}a_{ij}= begin{cases}
(-1)^{|i-j|}, & text{if} i neq j \
2, & text{if} i = j
end{cases}
end{equation}
Find $det(A)$. I was thinking to either form block matrices or do row operation, but it became a mess. Please help me to figure this out.

linear-algebra matrices determinant

share|cite|improve this question

edited Mar 12 at 5:13

Travis

63.3k768150

asked Mar 12 at 4:22

SutMarSutMar

1017

$endgroup$

add a comment | 

$begingroup$

I've been trying to solve this problem but I stucked.
Let $A = [a_{ij}]$ with size $ 2011 times 2011$ , and given the condition below

begin{equation}a_{ij}= begin{cases}
(-1)^{|i-j|}, & text{if} i neq j \
2, & text{if} i = j
end{cases}
end{equation}
Find $det(A)$. I was thinking to either form block matrices or do row operation, but it became a mess. Please help me to figure this out.

linear-algebra matrices determinant

share|cite|improve this question

edited Mar 12 at 5:13

Travis

63.3k768150

asked Mar 12 at 4:22

SutMarSutMar

1017

$endgroup$

share|cite|improve this question

edited Mar 12 at 5:13

Travis

63.3k768150

asked Mar 12 at 4:22

SutMarSutMar

1017

share|cite|improve this question

edited Mar 12 at 5:13

Travis

63.3k768150

asked Mar 12 at 4:22

SutMarSutMar

1017

edited Mar 12 at 5:13

Travis

63.3k768150

edited Mar 12 at 5:13

Travis

63.3k768150

asked Mar 12 at 4:22

SutMarSutMar

1017

asked Mar 12 at 4:22

SutMarSutMar

1017

1 Answer
1

active

oldest

votes

2

$begingroup$

Hint We can write $A$ as $$A = I_{2011} + {bf x} {bf x}^top ,$$ where ${bf x} = (1, -1, 1, -1, ldots, -1, 1)^top$. Now apply the Matrix Determinant Lemma.

share|cite|improve this answer

edited Mar 14 at 2:24

answered Mar 12 at 4:37

TravisTravis

63.3k768150

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks very much for the answer. I have found that I could do row operations and will leave 1's on the diagonal entries (1st to 2010-th) except the 2011-th entry. (the last diagonal yield "2012"). So, the determinant is just the product of diagonal entries = $1 overbrace{times 1 times 1 times cdot times 1}^{2010 times} times 2012 = 2012$
  $endgroup$
  – SutMar
  Mar 14 at 4:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yours is quite a nice argument, too!
  $endgroup$
  – Travis
  Mar 14 at 4:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (And you're welcome.)
  $endgroup$
  – Travis
  Mar 14 at 4:12
  

  
  
  
  

add a comment | 

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2

$begingroup$

Hint We can write $A$ as $$A = I_{2011} + {bf x} {bf x}^top ,$$ where ${bf x} = (1, -1, 1, -1, ldots, -1, 1)^top$. Now apply the Matrix Determinant Lemma.

share|cite|improve this answer

edited Mar 14 at 2:24

answered Mar 12 at 4:37

TravisTravis

63.3k768150

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks very much for the answer. I have found that I could do row operations and will leave 1's on the diagonal entries (1st to 2010-th) except the 2011-th entry. (the last diagonal yield "2012"). So, the determinant is just the product of diagonal entries = $1 overbrace{times 1 times 1 times cdot times 1}^{2010 times} times 2012 = 2012$
  $endgroup$
  – SutMar
  Mar 14 at 4:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yours is quite a nice argument, too!
  $endgroup$
  – Travis
  Mar 14 at 4:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (And you're welcome.)
  $endgroup$
  – Travis
  Mar 14 at 4:12
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint We can write $A$ as $$A = I_{2011} + {bf x} {bf x}^top ,$$ where ${bf x} = (1, -1, 1, -1, ldots, -1, 1)^top$. Now apply the Matrix Determinant Lemma.

share|cite|improve this answer

edited Mar 14 at 2:24

answered Mar 12 at 4:37

TravisTravis

63.3k768150

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks very much for the answer. I have found that I could do row operations and will leave 1's on the diagonal entries (1st to 2010-th) except the 2011-th entry. (the last diagonal yield "2012"). So, the determinant is just the product of diagonal entries = $1 overbrace{times 1 times 1 times cdot times 1}^{2010 times} times 2012 = 2012$
  $endgroup$
  – SutMar
  Mar 14 at 4:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yours is quite a nice argument, too!
  $endgroup$
  – Travis
  Mar 14 at 4:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (And you're welcome.)
  $endgroup$
  – Travis
  Mar 14 at 4:12
  

  
  
  
  

add a comment | 

$begingroup$

Hint We can write $A$ as $$A = I_{2011} + {bf x} {bf x}^top ,$$ where ${bf x} = (1, -1, 1, -1, ldots, -1, 1)^top$. Now apply the Matrix Determinant Lemma.

share|cite|improve this answer

edited Mar 14 at 2:24

answered Mar 12 at 4:37

TravisTravis

63.3k768150

$endgroup$

share|cite|improve this answer

edited Mar 14 at 2:24

answered Mar 12 at 4:37

TravisTravis

63.3k768150

answered Mar 12 at 4:37

TravisTravis

63.3k768150

answered Mar 12 at 4:37

TravisTravis

63.3k768150