Adjugate of Adjugate Matrix Determinant Problem

 

❓ Question

Let $A$ be a $3 \times 3$ matrix such that

$$\mathrm{\mid }\mathrm{adj}(\mathrm{adj}(\mathrm{adj}A))\mathrm{\mid }=81.$$

If

$$S=\{n\in \mathbb{Z}:\mid \mathrm{adj}(\mathrm{adj}A){\mid }^{\frac{(n-1{)}^2}{2}}=\mathrm{\mid }A{\mathrm{\mid }}^{3n^2-5n-4}\},$$

then

$$\sum _{n\in S}\mathrm{\mid }A{\mathrm{\mid }}^{n^2+n}$$

is equal to ?

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Determinant of adjoint matrix
👉 Power comparison
👉 Quadratic equations.

Main idea:

For an $n\times n$ matrix:

$$|\operatorname{adj}A|=|A|^{n-1}$$

Here matrix is $3\times3$.

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🔷 Step 1 — Find $|A|$ 💯

For $3\times3$ matrix:

$$|\operatorname{adj}A|=|A|^2$$

Now:

$$|\operatorname{adj}(\operatorname{adj}A)| = (|\operatorname{adj}A|)^2 = (|A|^2)^2 = |A|^4$$

Again:

$$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))| = (|A|^4)^2 = |A|^8$$

Given:

$$|A|^8=81$$
$$81=3^4$$

Thus:

$$|A|=3^{1/2}=\sqrt3$$

(Determinant magnitude positive for required powers.)

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🔷 Step 2 — Simplify Given Set Condition

Given:

$$\left| \operatorname{adj}(\operatorname{adj}A) \right|^{\frac{(n-1)^2}{2}} = |A|^{(3n^2-5n-4)}$$

But:

$$|\operatorname{adj}(\operatorname{adj}A)|=|A|^4$$

So:

$$(|A|^4)^{\frac{(n-1)^2}{2}} = |A|^{3n^2-5n-4}$$
$$|A|^{2(n-1)^2} = |A|^{3n^2-5n-4}$$

Since bases same and $|A|\neq1$,

equate powers:

$$2(n-1)^2=3n^2-5n-4$$

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🔷 Step 3 — Solve Quadratic

Expand:

$$2(n^2-2n+1)=3n^2-5n-4$$
$$2n^2-4n+2=3n^2-5n-4$$
$$0=n^2-n-6$$
$$(n-3)(n+2)=0$$

Thus:

$$\boxed{ n=3,\ -2 }$$

Hence:

$$S=\{3,-2\}$$

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🔷 Step 4 — Evaluate Required Sum

We need:

$$\sum_{n\in S}|A|^{(n^2+n)}$$

For $n=3$:

$$n^2+n=12$$
$$|A|^{12}=(\sqrt3)^{12}=3^6=729$$

For $n=-2$:

$$n^2+n=2$$
$$|A|^2=(\sqrt3)^2=3$$

Total:

$$729+3 = \boxed{ 732 }$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Using wrong formula:

$$|\operatorname{adj}A|=|A|^{n}$$

Correct formula:

$$\boxed{ |\operatorname{adj}A|=|A|^{n-1} }$$

For $3\times3$:

$$|\operatorname{adj}A|=|A|^2$$

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✅ Final Answer

$$\boxed{ 732 }$$

(Option 3)

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