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I am trying to calculate the determinant of the matrix:

$ \left(\begin{matrix} -6 & -5 & & \dots & -5\\ -5 & -6 & \dots & & -5 \\ \vdots & & \ddots & & \vdots \\ -5 & \dots & -5 & -6 & -5\\ -5 & \dots & & -5 & -6 \end{matrix}\right)\in \mathbb{R}^{82\times 82} $

How am I supposed to do this with a matrix so large? I tried looking for a pattern of Gauss Elimination but I got stuck. For clarification, the diagonal entries are $-6$ and the rest of the entries are $-5$.

4 Answers4

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The matrix has the form $-I - 5uu^T$ where $I$ is the $82\times 82$ identify matrix and $u$ is a $82 \times 1$ matrix with all entries one. There is a matrix determinant lemma for evaluating the determinant of this sort of matrix.

For any $n \times n$ invertible matrix $A$ and $n \times 1$ column matrix $u$, $v$, one has $$\det(A + uv^T) = \det(A)( 1 + v^TA^{-1}u)$$

Apply this to our matrix, we find

$$\det(-I - 5uu^T) = (-1)^{82} \det(I + 5uu^T) = 1 + 5u^Tu = 1 + 5\cdot 82 = 411$$

achille hui
  • 122,701
2

First, try to substract the third row from the second row. Then the third from the second... and so on. You will see a quite nice matrix (almost triangular and lots of zeros): $$\left(\begin{array}{cccccc} -6&-5&-5&-5&\dots&-5 \\ 0&-1&1&0&\dots&0\\ 0&0&-1&1&\dots&0\\ 0&0&0&-1&\dots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&0&0&0&\dots&-1 \end{array}\right)$$ Now, you can easily remove the "-5"s. Use the second row to obtain a zero in the second element of the first row. Then the third row for the third element and so on.

I hope you can continue from here.

AugSB
  • 5,007
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Let $M = \left(\begin{matrix} -6 & -5 & & \dots & -5\\ -5 & -6 & \dots & & -5 \\ \vdots & & \ddots & & \vdots \\ -5 & \dots & -5 & -6 & -5\\ -5 & \dots & & -5 & -6 \end{matrix}\right)$.

Then $M=\left(\begin{matrix} -5 & -5 & & \dots & -5\\ -5 & -5 & \dots & & -5 \\ \vdots & & \ddots & & \vdots \\ -5 & \dots & -5 & -5 & -5\\ -5 & \dots & & -5 & -5 \end{matrix}\right)+\left(\begin{matrix} -1 & 0 & & \dots & 0\\ 0 & -1 & \dots & & 0\\ \vdots & & \ddots & & \vdots \\ 0 & \dots & 0 & -1 & 0\\ 0 & \dots & & 0 & -1 \end{matrix}\right)$

$= \left(\begin{matrix} 5\\ 5\\ \vdots \\ 5\\ 5 \end{matrix}\right)\left(\begin{matrix} -1 & -1 & & \dots & -1\\ \end{matrix}\right)+\left(\begin{matrix} -1 & 0 & & \dots & 0\\ 0 & -1 & \dots & & 0\\ \vdots & & \ddots & & \vdots \\ 0 & \dots & 0 & -1 & 0\\ 0 & \dots & & 0 & -1 \end{matrix}\right)$.

By the matrix determinant lemma, $\det M=\left(1+\left(\begin{matrix} -1 & -1 & & \dots & -1\\ \end{matrix}\right)\left(\begin{matrix} -1 & 0 & & \dots & 0\\ 0 & -1 & \dots & & 0\\ \vdots & & \ddots & & \vdots \\ 0 & \dots & 0 & -1 & 0\\ 0 & \dots & & 0 & -1 \end{matrix}\right)\left(\begin{matrix} 5\\ 5\\ \vdots \\ 5\\ 5 \end{matrix}\right)\right)\cdot \left\vert\begin{matrix} -1 & 0 & & \dots & 0\\ 0 & -1 & \dots & & 0\\ \vdots & & \ddots & & \vdots \\ 0 & \dots & 0 & -1 & 0\\ 0 & \dots & & 0 & -1 \end{matrix}\right\vert=(1+82\cdot5)\cdot1=411.$

Steve Kass
  • 14,881
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This is not really an answer, just an advice.

Use some (free) software to check your results in cases like this.

I really like R, because it's (a) free, (b) great with vectors and matrices and (c) good for plotting.

Here's a little program to find the determinant:

N <- 82;
M <- matrix(rep(-5,N^2), nrow=N, ncol=N, byrow=TRUE);
D <- diag(rep(1,N));
det(M-D)

It gives $411$.

Just in case, I also checked with Mathematica which is also great, but definitely not free.

Yuriy S
  • 31,474