
#1
Jul612, 09:12 PM

P: 87

Let $$A$$ be the $$n\times n$$ matrix:
\begin{equation} A= \begin{bmatrix} % or pmatrix or bmatrix or Bmatrix or ... 2 & 1&\dots & 1 & 1 \\ 1 & 2&\dots & 1 & 1 \\ \vdots&\ddots & \ddots & 2 & 1 \\ 1 & & \dots & 1&2 \\ \end{bmatrix} \end{equation} (2s along the diagonal and 1s everywhere else.) Compute the determinant of $$A$$. Here is what I did. As usual, let $$I$$ be the identity matrix. Observe that $$A=I+B$$ where $$B$$ is a square matrix such that every element is 1. Next we can use the Taylor series expansion of the determinant \begin{align} \det(I + B) = \sum_{k=0}^{\infty} \frac{1}{k!} \left(  \sum_{j=1}^{\infty} \frac{(1)^j}{j}\mathrm{tr}(B^j) \right) ^k.\end{align} Note that $$\mathrm{tr}\left(B^j\right)$$ is always $$n^j$$, and so it follows that \begin{align} \det(A)& = \sum_{k=0}^{\infty} \frac{1}{k!} \left(  \sum_{j=1}^{\infty} \frac{(1)^j}{j}n^j \right) ^k\, , \end{align} The well known series expansions \begin{equation} \sum _{j=1}^{\infty }{\frac { \left( 1 \right) ^{j}}{j}}{n}^{j}=\log(n+1) \end{equation} and \begin{equation} \sum_{k=0}^\infty\frac{f(n)^k}{k!}=\exp{f(n)} \end{equation} allow us to conclude that the determinant is $$n+1.$$ I was wondering if there was a simpler way to do this problem. 



#2
Jul612, 09:45 PM

P: 606

1) It's easy to see, inductively, that [tex]\begin{equation}\left\begin{pmatrix} 1&1&1&....&1\\1&2&1&...&1\\...&...&...&...&...\\1&1&1&...&2\end{pmatrix }\right\end{equation}=1[/tex]Say, substract first row from second, develop by minors of the new 2nd row, etc. So substracting the third row from the 2nd one in the original matrix, we get: [tex]\begin{equation}\left\begin{matrix} 2&1&1&...&1\\1&2&1&...&1\\0&\!\!\!1&1&...&0\\...&...&...&...&...\\1&1&1&...&2 \end{matrix}\right\end{equation}[/tex] Developing wrt the third row, using the above fact and induction we get what we want. DonAntonio 



#3
Jul712, 11:54 AM

P: 87

Thanks DonAntonio




#4
Jul712, 06:23 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,386

How to compute the determinant of this matrix?
Another way is to factorize A = LU where L is unit lower triangular and U is upper triangular. Use Crout's verison of Gaussian elimination.
The diagonals of U are 2, 3/2, 4/3, 5/4, .... (n+1)/n. The determinant is the product of the diagonals = n+1 



#5
Jul812, 12:08 AM

P: 87

Thanks AlephZero, I am going to learn about Crouts version.



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