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Let $A=[a_{ij}]\in Mat_n(\mathbb{R})$ the matrix defined by:

$$a_{ij}=\begin{eqnarray}n+1 &&&&&&&&\text{if $i=j$}\\ 1 &&&&&&&&\text{if $i\not=j$} \end{eqnarray}$$

Calulate $\det(A)$

My work:

By definition $$A=\begin{pmatrix} n+1&&1&&1&&1&&...&&1\\ 1&&n+1&&1&&1&&...&&1\\ 1&&1&&n+1&&1&&...&&1\\ .\\ .\\ .\\ 1&&1&&1&&1&&...&&n+1 \end{pmatrix}$$

Apply elemental operation by rows, we have: $$\begin{pmatrix} n+1&&1&&1&&1&&...&&1\\ 1&&n+1&&1&&1&&...&&1\\ 1&&1&&n+1&&1&&...&&1\\ .\\ .\\ .\\ 1&&1&&1&&1&&...&&n+1 \end{pmatrix}\equiv \begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 1&&n+1&&1&&1&&...&&1\\ 1&&1&&n+1&&1&&...&&1\\ .\\ .\\ .\\ n+1&&1&&1&&1&&...&&1 \end{pmatrix}\\\\ \equiv \begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 0&&n&&0&&0&&...&&-n\\ 0&&0&&n&&0&&...&&-n\\ .\\ .\\ .\\ n&&0&&0&&0&&...&&-n \end{pmatrix} $$ $$ \equiv \begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 0&&n&&0&&0&&...&&-n\\ 0&&0&&n&&0&&...&&-n\\ .\\ .\\ .\\ 0&&-n&&-n&&-n&&...&&-n^2-2n \end{pmatrix} \equiv \begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 0&&n&&0&&0&&...&&-n\\ 0&&0&&n&&0&&...&&-n\\ .\\ .\\ .\\ 0&&0&&-n&&-n&&...&&-n^2-3n \end{pmatrix} \equiv \begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 0&&n&&0&&0&&...&&-n\\ 0&&0&&n&&0&&...&&-n\\ .\\ .\\ .\\ 0&&0&&0&&-n&&...&&-n^2-4n \end{pmatrix} \equiv \begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 0&&n&&0&&0&&...&&-n\\ 0&&0&&n&&0&&...&&-n\\ .\\ .\\ .\\ 0&&0&&0&&0&&...&&n^2-n^n \end{pmatrix}=B $$

Then, $\det(A)=-\det(B)=-\det\begin{pmatrix} 1&&1&&1&&1&&...&&n+1\\ 0&&n&&0&&0&&...&&-n\\ 0&&0&&n&&0&&...&&-n\\ .\\ .\\ .\\ 0&&0&&0&&0&&...&&n^2-n^n \end{pmatrix}=-n^{n-1}+n^n$

I have a little doubt in the $n\times n$ element, I think is a little different. Can someone help me?

carmichael561
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rcoder
  • 4,665

2 Answers2

2

Note that $A=nI_n+vv^T$, where $v=(1,\dots,1)^T$. Using the matrix determinant lemma, we have $$ \det(nI_n+vv^T)=\Big(1+\frac{1}{n}v^Tv\Big)\det(nI_n)=2n^n$$

carmichael561
  • 54,793
1

Writing $A=nI_n+U$, where $U$ is the matrix with all elements equal to one, the eigenvalues $\lambda$ are solutions to the characteristic polynomial $\det(U-(\lambda-n)I_n)=\det(U-\lambda'I_n)$, where $\lambda'=\lambda-n$. In this question was shown that $U$ has all eigenvalues $\lambda'$ equal to $0$, except for one that is equal to $n$. Then $A$ has all eigenvalues $\lambda$ equal to $n$, except for one that is equal to $2n$. As the determinant is the product of eigenvalues, $\det(A)=n^{n-1}\cdot 2n=2n^n$.

Joca Ramiro
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