Repeated adjugate determinant modulo seven

Key concept: This matrix problem uses the determinant identity for adjugates and modular exponent cycling.

Step by step

Step 1: Use the determinant identity for a $3\times 3$ matrix

For an $n\times n$ matrix $A$,

$\det(\operatorname{adj}(A))=\det(A)^{n-1}.$

Here $n=3$ and $\det(A)=3$, so

$\det(\operatorname{adj}(A))=3^{2}=9.$

Let

$d_k=\det(\operatorname{adj}^{(k)}(A)),$

where $\operatorname{adj}^{(k)}$ means applying $\operatorname{adj}$ $k$ times.

Then for a $3\times 3$ matrix,

$d_{k+1}=d_k^2.$

So the sequence begins

$d_0=3,\quad d_1=3^2,\quad d_2=3^4,\quad d_3=3^8,$

and in general

$d_k=3^{2^k}.$

Step 2: Reduce the exponent modulo 6

We need

$m=3^{2^{2023}}\pmod 7.$

Because 7 is prime and 3 is not divisible by 7,

$3^6\equiv 1\pmod 7.$

So we only need the exponent modulo 6.

Now compute

$2^{2023}\pmod 6.$

For powers of 2 modulo 6:

$2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 2,\quad 2^4\equiv 4,$

so the pattern alternates between 2 and 4. Since 2023 is odd,

$2^{2023}\equiv 2\pmod 6.$

Therefore,

$m\equiv 3^2\equiv 9\equiv 2\pmod 7.$

Step 3: Final answer

The remainder when $m$ is divided by 7 is

$\boxed{2}.$

Why the pattern is so fast

Repeated adjugation on a $3\times 3$ matrix squares the determinant each time, so the exponent doubles every step. That makes the determinant grow as a power tower in the exponent, which is exactly why modular arithmetic is the right tool here.

Pitfall alert

A common mistake is to think the determinant of the adjugate stays the same under repeated application. It does not: for a $3\times 3$ matrix, each adjugate application squares the determinant. Another frequent error is trying to compute $3^{2^{2023}}$ directly rather than using Fermat’s little theorem and the cycle of exponents modulo 6. The shortcut depends on reducing the exponent first, not the base.

Try different conditions

If the matrix were $4\times 4$ with $\det(A)=3$, the identity would change to

$\det(\operatorname{adj}(A))=\det(A)^3,$

so repeated adjugation would no longer square the determinant each time. Instead, each step would cube the current determinant. The modular strategy would still be the same, but the exponent growth and the period modulo 7 would be different. You would first derive the new recurrence before reducing modulo 7.

Further reading

adjugate matrix, determinant identity, modular exponentiation