JEE Mathematics 2023 Question with Solution

Given: $P^T = aP + (a-1)I$ for a $3 \times 3$ real matrix $P$, with $a > 1$.

Find: the correct statement about $P$ and $|\operatorname{Adj} P|$.

From the given relation,

$$P^T = aP + (a-1)I$$

Taking transpose again as used in the solution,

$$P = aP^T + (a-1)I$$

Now subtract the two relations appropriately:

$$P^T - P = a(P - P^T)$$

Hence,

$$(1+a)(P^T - P) = 0$$

Since $a > 1$, we have $a \ne -1$. Therefore,

$$P^T = P$$

So $P$ is symmetric.

Substituting $P^T = P$ into the original equation,

$$P = aP + (a-1)I$$

Thus,

$$(1-a)P = (a-1)I$$

which gives

$$P = -I$$

Therefore,

$$|P| = |-I| = (-1)^3 = -1$$

so $|P| \ne 0$, hence $P$ is not singular.

For a $3 \times 3$ matrix,

$$|\operatorname{Adj} P| = |P|^{3-1} = |P|^2$$

Therefore,

$$|\operatorname{Adj} P| = (-1)^2 = 1$$

So, the correct option is D.

After obtaining $P = -I$ from the given relation, compute the determinant first:

$$|P| = |-I| = -1$$

Now use the standard result for an $n \times n$ matrix:

$$|\operatorname{Adj} P| = |P|^{n-1}$$

Here $n = 3$, so

$$|\operatorname{Adj} P| = |P|^2 = (-1)^2 = 1$$

Hence the correct statement is $|\operatorname{Adj} P| = 1$.