# How to compute the determinant of this matrix?

by Charles49
Tags: compute, determinant, matrix
 P: 87 Let $$A$$ be the $$n\times n$$ matrix: 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} (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 -\sum _{j=1}^{\infty }{\frac { \left( -1 \right) ^{j}}{j}}{n}^{j}=\log(n+1) and \sum_{k=0}^\infty\frac{f(n)^k}{k!}=\exp{f(n)} allow us to conclude that the determinant is $$n+1.$$ I was wondering if there was a simpler way to do this problem.
P: 606
 Quote by Charles49 Let $$A$$ be the $$n\times n$$ matrix: 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} (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 -\sum _{j=1}^{\infty }{\frac { \left( -1 \right) ^{j}}{j}}{n}^{j}=\log(n+1) and \sum_{k=0}^\infty\frac{f(n)^k}{k!}=\exp{f(n)} allow us to conclude that the determinant is $$n+1.$$ I was wondering if there was a simpler way to do this problem.

1) It's easy to see, inductively, that
$$$$\left|\begin{pmatrix} 1&1&1&....&1\\1&2&1&...&1\\...&...&...&...&...\\1&1&1&...&2\end{pmatrix }\right|$$=1$$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:

$$$$\left|\begin{matrix} 2&1&1&...&1\\1&2&1&...&1\\0&\!\!\!-1&1&...&0\\...&...&...&...&...\\1&1&1&...&2 \end{matrix}\right|$$$$
Developing wrt the third row, using the above fact and induction we get what we want.

DonAntonio
 P: 87 Thanks DonAntonio
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