Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj} (\operatorname{adj}(\operatorname{adj} A))| = 81$. If $$S = \left\{ n \in \mathbb{Z} : |\operatorname{adj}(\operatorname{adj} A)|^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4} \right\},$$ then $\sum_{n \in S} \left| A^{n^2 + n} \right|$ is equal to:

Step 1: Determine the determinant of A

For a $k \times k$ matrix $A$, the determinant of the adjoint of the adjoint of the adjoint of $A$ is given by the formula $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = |A|^{(k-1)^3}$. In this problem, $A$ is a $3 \times 3$ matrix, so $k=3$. We are given that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 81$. We can use this to find the value of $|A|$.

Step 2: Calculate |A|

Substituting $k=3$ into the formula, we get $|A|^{(3-1)^3} = |A|^8$. We are given that this is equal to 81. Since $81 = 3^4$, we have $|A|^8 = 3^4$. Taking the eighth root of both sides, we find that $|A| = \pm (3^4)^{1/8} = \pm 3^{1/2} = \pm \sqrt{3}$.

Step 3: Simplify the expression for |adj(adj A)|

For a $k \times k$ matrix $A$, the determinant of the adjoint of the adjoint of $A$ is given by the formula $|\operatorname{adj}(\operatorname{adj} A)| = |A|^{(k-1)^2}$. Since $k=3$, this simplifies to $|A|^{(3-1)^2} = |A|^4$.

Step 4: Substitute into the given equation for S

Now we substitute $|\operatorname{adj}(\operatorname{adj} A)| = |A|^4$ into the given equation for set $S$. This gives us $(|A|^4)^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$. Using the exponent rule $(a^b)^c = a^{bc}$, the left side becomes $|A|^{4 \cdot \frac{(n-1)^2}{2}} = |A|^{2(n-1)^2}$.

Step 5: Solve for n

Since $|A| = \pm \sqrt{3}$ and $|A| \neq 0, 1$, we can equate the exponents: $2(n-1)^2 = 3n^2 - 5n - 4$. Expanding and simplifying the quadratic equation: $2(n^2 - 2n + 1) = 3n^2 - 5n - 4 \implies 2n^2 - 4n + 2 = 3n^2 - 5n - 4$. Rearranging terms gives $n^2 - n - 6 = 0$. Factoring the quadratic equation yields $(n-3)(n+2) = 0$. Thus, the possible values for $n$ are $n=3$ and $n=-2$. So, $S = \{-2, 3\}$.

Step 6: Calculate the sum of the determinants

We need to calculate $\sum_{n \in S} |A^{n^2 + n}|$. We know that $|A^m| = |A|^m$. So, we need to calculate $|A|^{n^2+n}$ for $n=3$ and $n=-2$. For $n=3$, the exponent is $3^2+3 = 9+3 = 12$. So, $|A|^{12} = (\pm \sqrt{3})^{12} = (3^{1/2})^{12} = 3^6 = 729$. For $n=-2$, the exponent is $(-2)^2+(-2) = 4-2 = 2$. So, $|A|^2 = (\pm \sqrt{3})^2 = 3$. The sum is $729 + 3 = 732$.

More from this chapter

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