JEE Mathematics 2023 Question with Solution

Given: $A$ is a $3 \times 3$ matrix and $|adj(adj(A))| = 12^4$.

Find: $|A^{-1}adj(A)|$.

Using the determinant property for an $n \times n$ matrix,

$$|adj(A)| = |A|^{n-1}$$

Hence,

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

For $n = 3$,

$$|adj(adj(A))| = |A|^4$$

So,

$$|A|^4 = 12^4$$

Therefore,

$$|A| = 12$$

Now,

$$|A^{-1}adj(A)| = |A^{-1}|\,|adj(A)|$$

Also,

$$|A^{-1}| = \frac{1}{|A|}, \qquad |adj(A)| = |A|^{3-1} = |A|^2$$

Thus,

$$|A^{-1}adj(A)| = \frac{1}{|A|}\cdot |A|^2 = |A| = 12$$

Therefore, the correct option from the extracted solution working is C. The solution contains an internal inconsistency because it also states $2\sqrt{3}$, but the determinant relation shown implies $12$.

The solution text uses the step

$$|A|^{(n-1)^3} = 12^4$$

and with $n=3$ concludes

$$|A|^8 = 12^4$$

which gives

$$|A|^2 = 12 \Rightarrow |A| = 2\sqrt{3}$$

This is the route used there to obtain $|A^{-1}adj(A)| = 2\sqrt{3}$.

However, for a $3 \times 3$ matrix,

$$|adj(A)| = |A|^{2}$$

and therefore

$$|adj(adj(A))| = |adj(A)|^2 = |A|^4$$

not $|A|^8$.

So the worked algebra displayed on the page is inconsistent with the standard determinant identity. Using the identity written correctly gives

$$|A^{-1}adj(A)| = 12$$

which matches option C.