How to compute the determinant of this matrix?

  1. 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. jcsd


  3. 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
     
  4. Thanks DonAntonio
     
  5. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

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