JEE Mathematics 2023 Question with Solution

Given: $|A| = 2$ and

$$\left|\text{Adj}(2\,\text{Adj}(2A^{-1}))\right| = 2^{84}$$

Find: $n$

Use the property for an $n \times n$ matrix $M$:

$$|\text{Adj}(M)| = |M|^{n-1}$$

Applying this to $M = 2\,\text{Adj}(2A^{-1})$,

$$\left|\text{Adj}(2\,\text{Adj}(2A^{-1}))\right| = \left|2\,\text{Adj}(2A^{-1})\right|^{n-1}$$

Now,

$$\left|2\,\text{Adj}(2A^{-1})\right| = 2^n\left|\text{Adj}(2A^{-1})\right|$$

Also,

$$\left|\text{Adj}(2A^{-1})\right| = |2A^{-1}|^{n-1}$$

Since

$$|2A^{-1}| = 2^n|A^{-1}| = 2^n \cdot \frac{1}{|A|} = 2^n \cdot \frac{1}{2} = 2^{n-1}$$

we get

$$\left|\text{Adj}(2A^{-1})\right| = \left(2^{n-1}\right)^{n-1} = 2^{(n-1)^2}$$

Therefore,

$$\left|2\,\text{Adj}(2A^{-1})\right| = 2^n \cdot 2^{(n-1)^2} = 2^{n + (n-1)^2}$$

Hence,

$$\left|\text{Adj}(2\,\text{Adj}(2A^{-1}))\right| = \left(2^{n + (n-1)^2}\right)^{n-1} = 2^{(n-1)(n + (n-1)^2)}$$

Given this equals $2^{84}$,

$$(n-1)(n + (n-1)^2) = 84$$ $$(n-1)(n^2 - n + 1) = 84$$

Trying $n = 5$,

$$(5-1)(25-5+1) = 4 \cdot 21 = 84$$

which satisfies the equation.

Therefore, the value of $n$ is $5$.

Given: $|A| = 2$

A common error is to use

$$|\text{Adj}(M)| = 2^{n-1}|M|$$

which is incorrect. The correct identity is

$$|\text{Adj}(M)| = |M|^{n-1}$$

for any $n \times n$ matrix $M$.

Let

$$M = 2\,\text{Adj}(2A^{-1})$$

Then

$$|\text{Adj}(M)| = |M|^{n-1}$$

Now evaluate $|M|$:

$$|M| = \left|2\,\text{Adj}(2A^{-1})\right| = 2^n\left|\text{Adj}(2A^{-1})\right|$$

Again,

$$\left|\text{Adj}(2A^{-1})\right| = |2A^{-1}|^{n-1}$$

And

$$|2A^{-1}| = 2^n|A^{-1}| = 2^n \cdot \frac{1}{2} = 2^{n-1}$$

So,

$$\left|\text{Adj}(2A^{-1})\right| = 2^{(n-1)^2}$$

Thus,

$$|M| = 2^{n + (n-1)^2}$$

Therefore,

$$|\text{Adj}(M)| = 2^{(n-1)(n + (n-1)^2)}$$

Comparing with $2^{84}$,

$$(n-1)(n^2 - n + 1) = 84$$

This gives $n = 5$.

So the correct answer is $5$. The alternate provided approach concluding $85$ is inconsistent with its own final statement and with the determinant property, so the correct value remains $5$.