Dtjytyk

How can a new country break out from a developed country without war?

Is VPN a layer 3 concept?

Print last inputted byte

Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run?

Can other pieces capture a threatening piece and prevent a checkmate?

Would mining huge amounts of resources on the Moon change its orbit?

How to test the sharpness of a knife?

How can an organ that provides biological immortality be unable to regenerate?

PTIJ: Which Dr. Seuss books should one obtain?

Error in master's thesis, I do not know what to do

Is there any common country to visit for uk and schengen visa?

How do you justify more code being written by following clean code practices?

Should a narrator ever describe things based on a characters view instead of fact?

Norwegian Refugee travel document

How are passwords stolen from companies if they only store hashes?

Why is this tree refusing to shed its dead leaves?

Does fire aspect on a sword, destroy mob drops?

Why didn’t Eve recognize the little cockroach as a living organism?

Which partition to make active?

Knife as defense against stray dogs

Have any astronauts/cosmonauts died in space?

Homology of the fiber

Was World War I a war of liberals against authoritarians?

Help with identifying unique aircraft over NE Pennsylvania

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 | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3144643%2fdeterminant-of-a-particular-large-matrix%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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