Today's number, \( \textbf{27996} \) that defines my diurnal age has an interesting property that qualifies it for membership in OEIS A097401: Largest achievable determinant of a 3 x 3 matrix whose elements are 9 distinct nonnegative integers chosen from the range \( 0 \dots n\). In the case of 27996, \(n = 27\). The OEIS comments state that the configuration that achieves this largest value looks like this:$$ \begin{bmatrix} n & n-5 & 1 \\ 2 & n-1 & n-3 \\ n-4 & 0 & n-2 \end{bmatrix}$$The associated determinant is given by:$$ \begin{vmatrix} n & n-5 & 1 \\ 2 & n-1 & n-3 \\ n-4 & 0 & n-2 \end{vmatrix} = 2(n^3 - 9n^2 + 34n - 42) $$So in the case of \(n=27\) we have (permalink):$$ \begin{align} \begin{bmatrix} 27 & 22 & 1 \\ 2 & 26 & 24 \\ 23 & 0 & 25 \end{bmatrix} \rightarrow \begin{vmatrix} 27 & 22 & 1 \\ 2 & 26 & 24 \\ 23 & 0 & 25 \end{vmatrix}=27996 \end{align} $$Now the minimal value of \(n\) must be 8 in order to accommodate the nine digits from 0 to 8. This gives the following configuration for the maximum value of the determinant and if we add 1 to each entry we get the configuration for the digits from 1 to 9:$$ \begin{align} \begin{bmatrix} 8 & 3 & 1 \\ 2 & 7 & 5 \\ 4 & 0 & 6\end{bmatrix} \rightarrow \begin{vmatrix} 8 & 3 & 1 \\ 2 & 7 & 5 \\ 4 & 0 & 6 \end{vmatrix}=332 \\ \\ \begin{bmatrix} 9 & 4 & 2 \\ 3 & 8 & 6 \\ 5 & 1 & 7\end{bmatrix} \rightarrow \begin{vmatrix} 9 & 4 & 2 \\ 3 & 8 & 6 \\ 5 & 1 & 7 \end{vmatrix}=412 \end{align}$$Minimum values are easily determined because there is a fundamental property of matrices stating that swapping any two rows (or columns) reverses the sign of the determinant. So the minimal values of the determinants are just the negative values of the positive determinants. The members of OEIS A097401 are:
332, 528, 796, 1148, 1596, 2152, 2828, 3636, 4588, 5696, 6972, 8428, 10076, 11928, 13996, 16292, 18828, 21616, 24668, 27996, 31612, 35528, 39756, 44308, 49196, 54432, 60028, 65996, 72348, 79096, 86252, 93828, 101836, 110288, 119196, 128572
Gemini gives the following explanation of how this maximal configuration is arrived at:
The configuration of a $3 \times 3$ matrix using the digits 1 through 9 that yields the maximum determinant results in a value of 412.
Here is the optimal matrix configuration:
$$\begin{bmatrix} 9 & 4 & 2 \\ 3 & 8 & 6 \\ 5 & 1 & 7 \end{bmatrix}$$How to Calculate the Determinant
To understand why this specific arrangement works, we can calculate the determinant using the Rule of Sarrus, which involves summing the products of the three downward diagonals and subtracting the products of the three upward diagonals.
1. The Positive Diagonals (Maximize these):
$9 \times 8 \times 7 = \mathbf{504}$
$4 \times 6 \times 5 = \mathbf{120}$
$2 \times 3 \times 1 = \mathbf{6}$
Sum of positives: $504 + 120 + 6 = \mathbf{630}$
2. The Negative Diagonals (Minimize these):
$2 \times 8 \times 5 = \mathbf{80}$
$4 \times 3 \times 7 = \mathbf{84}$
$9 \times 6 \times 1 = \mathbf{54}$
Sum of negatives: $80 + 84 + 54 = \mathbf{218}$
3. The Final Result:
The Strategy Behind the Solution
To achieve this maximum value, you cannot simply place numbers randomly. You must use a Grouping Strategy to exploit the math behind determinants:
Group the Numbers: Divide the digits into three tiers:
Large: $\{9, 8, 7\}$
Medium: $\{6, 5, 4\}$
Small: $\{3, 2, 1\}$
Maximize Positive Diagonals: You want the largest numbers to multiply each other.
We placed the Large group $(9, 8, 7)$ on the main diagonal to get the massive product of $504$.
We placed the Medium group $(6, 5, 4)$ on the next positive diagonal.
We sacrificed the Small group $(3, 2, 1)$ on the final positive diagonal.
Minimize Negative Diagonals: You want to avoid large numbers multiplying each other in the reverse direction. By carefully rotating the positions of the Medium and Small numbers, we ensured that every negative diagonal consists of a "mixed" product (one Large, one Medium, and one Small number), preventing any single negative term from becoming too large.
| Goal | Determinant Value | Strategy |
| Maximum | 412 | Maximize main diagonal products; minimize off-diagonal products. |
| Minimum | -412 | Swap two rows of the Maximum matrix. |
| Smallest Magnitude | 0 | Use rows that are linearly dependent (e.g., $1,2,3$ and $4,5,6$ and $7,8,9$). |
If you specifically require a non-zero (non-singular) result, the minimum possible magnitude is 1.
Here are the configurations for both scenarios:
1. The Singular Solution (Determinant = 0)
This is the absolute closest you can get to zero.
Why this happens:
In this matrix, the rows form an arithmetic progression (each number increases by 1).
Row 2 is just Row 1 with 3 added to every element.
Row 3 is just Row 2 with 3 added to every element.
In linear algebra terms, these rows are linearly dependent. If you subtract Row 1 from Row 2, and Row 2 from Row 3, you get identical rows of $(3, 3, 3)$, which collapses the determinant to zero.
2. The Non-Singular Solution (Determinant $\approx 0$)
If you need the matrix to be invertible (non-zero determinant), you want to break the perfect symmetry of the zero-matrix just enough to get a value of 1 or -1 (or very close to it).
Here is a configuration that yields a determinant of -2:
Calculation:
$1(36 - 48) - 2(27 - 42) + 5(24 - 28)$
$1(-12) - 2(-15) + 5(-4)$
$-12 + 30 - 20 = \mathbf{-2}$
(Note: It is possible to find arrangements that result in 1, but they are rare "needles in a haystack" among the 362,880 possible permutations. The value of -2 is an excellent example of a near-singular matrix.)
Summary Table
| Goal | Determinant | Configuration Strategy |
| Absolute Closest | 0 | Use sequential numbers (1,2,3 / 4,5,6...) to create linear dependence. |
| Smallest Non-Zero | $\pm$ 1 or 2 | Use the sequential matrix but swap one or two "Small" numbers with "Medium" ones to slightly break the pattern |
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