JEE Matrix Trick: adj(adj(adj A)) = 81? Solve This Determinant Puzzle! 🔥

 

❓ 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 Solution

We use two standard facts for an invertible $n \times n$ matrix $A$:

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

2. For 3×3, $n = 3 \Rightarrow \det(\operatorname{adj}A) = |A|^2$
   

From $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 81$, we express everything in terms of $|A|$, then solve the given equation for integers $n$. Finally we compute $\sum_{n\in S} |A|^{n^2+n}$

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🔹 Step 1 — Express all determinants in terms of $|A|$

Let

$$\mathrm{\mid }A\mathrm{\mid }=D\mathrm{.}$$

For a 3×3 matrix:

• First adjoint:

  $$B_1=\mathrm{adj}A,\mathrm{\mid }{B}_1\mathrm{\mid }=\mathrm{\mid }A{\mathrm{\mid }}^2=D^2\mathrm{.}$$

• Second adjoint:

  $$B_2=\mathrm{adj}(B_1),\mathrm{\mid }{B}_2\mathrm{\mid }=\mathrm{\mid }{B}_1{\mathrm{\mid }}^2=(D^2{)}^2=D^4\mathrm{.}$$

• Third adjoint:

  $$B_3=\mathrm{adj}(B_2),\mathrm{\mid }{B}_3\mathrm{\mid }=\mathrm{\mid }{B}_2{\mathrm{\mid }}^2=(D^4{)}^2=D^8\mathrm{.}$$

Given:

$$\mathrm{\mid }{B}_3\mathrm{\mid }=\mathrm{\mid }\mathrm{adj}(\mathrm{adj}(\mathrm{adj}A))\mathrm{\mid }=81,$$

so

$$D^8=81=3^4\mathrm{.}$$

Thus

$$\mathrm{\mid }A\mathrm{\mid }=D=\pm 3^{1\mathrm{/}2}=\pm \sqrt{3}\mathrm{.}$$

Also

$$\mid \mathrm{adj}(\mathrm{adj}A)\mid =\mathrm{\mid }{B}_2\mathrm{\mid }=D^4=(\sqrt{3}{)}^4=9.$$

So,

$$\mid \mathrm{adj}(\mathrm{adj}A)\mid =9.$$

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🔹 Step 2 — Use the condition defining S

We’re given:

$$\mid \mathrm{adj}(\mathrm{adj}A){\mid }^{\frac{(n-1{)}^2}{2}}=\mathrm{\mid }A{\mathrm{\mid }}^{3n^2-5n-4}\mathrm{.}$$

Substitute:

$$9^{\frac{(n-1{)}^2}{2}}=D^{3n^2-5n-4}\mathrm{.}$$

Write both sides as powers of 3.
Since $9 = 3^2$

$$9^{\frac{(n-1{)}^2}{2}}=(3^2{)}^{\frac{(n-1{)}^2}{2}}=3^{(n-1{)}^2}\mathrm{.}$$

Also $D^2 = 3$, and $D = \pm 3^{1/2}$. For any such $D$:

$$D^{3n^2-5n-4}=(3^{1\mathrm{/}2}{)}^{3n^2-5n-4}\text{ (up to a sign, which forces exponent even)}=3^{\frac{3n^2-5n-4}{2}}\mathrm{.}$$

So we equate exponents:

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

Multiply by 2:

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

Solve:

$$n^2 - n - 6 = 0 \Rightarrow (n-3)(n+2) = 0,$$

so

$$n = 3 \quad \text{or} \quad n = -2.$$

Thus:

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

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🔹 Step 3 — Compute $\displaystyle \sum_{n \in S} |A|^{\,n^2 + n}$

We need:

$$\sum_{n \in S} |A|^{\,n^2 + n} = |A|^{(-2)^2 + (-2)} + |A|^{3^2 + 3}.$$

Exponents:

• For $n = -2$:

  $$n^2+n=4-2=2.$$

• For $n = 3$:

  $$n^2+n=9+3=12.$$

So:

$$|A|^2 = D^2 = 3,$$
$$\mathrm{\mid }A{\mathrm{\mid }}^{12}=D^{12}=(D^2{)}^6=3^6=729.$$

Therefore:

$$\sum _{n\in S}\mathrm{\mid }A{\mathrm{\mid }}^{n^2+n}=3+729=732.$$

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✅ Final Answer

$$\boxed{732}$$

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