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Find the determinant of order $100$:

$$D=\begin{vmatrix} 5 &5 &5 &\ldots &5 &5 &-1\\ 5 &5 &5 &\ldots &5 &-1 &5\\ 5 &5 &5 &\ldots &-1 &5 &5\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 5 &5 &-1 &\ldots &5 &5 &5\\ 5 &-1 &5 &\ldots &5 &5 &5\\ -1 &5 &5 &\ldots &5 &5 &5 \end{vmatrix}$$

I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:

Multiplying the first row by $(-1)$ and adding it to all rows: $$D=\begin{vmatrix} 5 &5 &5 &\ldots &5 &5 &-1\\ 5 &5 &5 &\ldots &5 &-1 &5\\ 5 &5 &5 &\ldots &-1 &5 &5\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 5 &5 &-1 &\ldots &5 &5 &5\\ 5 &-1 &5 &\ldots &5 &5 &5\\ -1 &5 &5 &\ldots &5 &5 &5 \end{vmatrix}=\begin{vmatrix} 5 &5 &5 &\ldots &5 &5 &-1\\ 0 &0 &0 &\ldots &0 &-6 &6\\ 0 &0 &0 &\ldots &-6 &0 &6\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 0 &0 &-6 &\ldots &0 &0 &6\\\ 0 &-6 &0 &\ldots &0 &0 &6\\ -6 &0 &0 &\ldots &0 &0 &6 \end{vmatrix}$$

Applying Laplace's method to the first column

$$D=5\begin{vmatrix} 0 &0 &\ldots &0 &-6 &6\\ 0 &0 &\ldots &-6 &0 &6\\ \vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 0 &-6 &\ldots &0 &0 &6\\ -6 &0 &\ldots &0 &0 &6 \end{vmatrix}+6\begin{vmatrix} 5 &5 &\ldots &5 &5 &-1\\ 0 &0 &\ldots &0 &-6 &6\\ 0 &0 &\ldots &-6 &0 &6\\ \vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 0 &-6 &\ldots &0 &0 &6\\ -6 &0 &\ldots &0 &0 &6 \end{vmatrix}$$

I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?

lmc
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5 Answers5

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Using this matrix determinant lemma I am getting $(1+5\cdot100\cdot(-\frac{1}{6}))(-6)^{100}$.

Note about general size, one needs to think about the determinant of anti-diagonal matrix to compute it. My guess is that it equals $(1+5n(-\frac{1}{6}))(-1)^{\frac{n(n-1)}{2}}(-6)^{n}$.

Jan
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Add all the rows to the last row and in the last row all entries will be $99 \cdot 5 - 1 = 494$. Now you can take $494$ outside of the determinant and now start subtracting $-5$ times the last row from each other row. Then the entries in the second diagonal will be $-6$, the last row will be $1$'s , while all other entries will be $0$. Then you can start changing the first row with the last one, the second with the second to last and so on. As there will be $50$ changes (even number), the value of the determinant will not change. Eventually you will get a triangular number and hence the determinant is:

$$494 \cdot (-6)^{99}$$

hmakholm left over Monica
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Stefan4024
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If we reverse the order of the rows, we end up with a matrix of the form $M = A-6I$, where $A$ has $5$ for every entry. This requires $50$ transpositions, so the determinant of this new matrix is the same. To find this determinant, we can multiply the eigenvalues of the matrix $A - 6I$, which can be found to be $$ \{\overbrace{-6,-6,\dots,-6}^{99 \text{ times}},494\} $$ in the method outlined here.

Community
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Ben Grossmann
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Starting from $$D=\begin{vmatrix} 5 &5 &5 &\ldots &5 &5 &-1\\ 5 &5 &5 &\ldots &5 &-1 &5\\ 5 &5 &5 &\ldots &-1 &5 &5\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 5 &5 &-1 &\ldots &5 &5 &5\\ 5 &-1 &5 &\ldots &5 &5 &5\\ -1 &5 &5 &\ldots &5 &5 &5 \end{vmatrix}=\begin{vmatrix} 5 &5 &5 &\ldots &5 &5 &-1\\ 0 &0 &0 &\ldots &0 &-6 &6\\ 0 &0 &0 &\ldots &-6 &0 &6\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 0 &0 &-6 &\ldots &0 &0 &6\\\ 0 &-6 &0 &\ldots &0 &0 &6\\ -6 &0 &0 &\ldots &0 &0 &6 \end{vmatrix}$$ you can add each column to the last column to obtain $$ D=\begin{vmatrix} 5 &5 &5 &\ldots &5 &5 &-1+99\times5\\ 0 &0 &0 &\ldots &0 &-6 &0\\ 0 &0 &0 &\ldots &-6 &0 &0\\ \vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\ 0 &0 &-6 &\ldots &0 &0 &0\\\ 0 &-6 &0 &\ldots &0 &0 &0\\ -6 &0 &0 &\ldots &0 &0 &0 \end{vmatrix}. $$

Now the matrix is upper-tridiagonal (along the not so conventional diagonal I have to admit). By Leibniz formula (or Laplace along the last column), it can be seen that the determinant is the product of the diagonal elements. This yields $$D= (-1+99\times5)(-6)^{99} =-494\times 6^{99}.$$

Fabian
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you start with a $2×2$ matrix and see the eigenvalues they are $4,6$. Then see for $3×3$ matrix the eigenvalues are $9,6,-6$ for $4×4$ they are $14,6,-6,6.$. Hence whenever the order is even the eigenvalues 6 exceeds the eigenvalue $-6$ by 1 in multiplicity. and hence for even $n$ the snswer is

$det= (5(n-1)-1).6^{n/2}.(-6)^{\frac{n}{2}-1}$ In your case the answer is $494.(6)^{50}.(-6)^{49}$

Upstart
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