JEE Mathematics 2023 Question with Solution

Given: $Q = PAP^T$, where $P$ is orthogonal, so $P^TP = I$. We need $P^TQ^{2007}P = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

Find: The value of $2a + b - 3c - 4d$.

Using similarity transformation,

Therefore,

Now compute powers of

A standard pattern gives

Hence,

So,

Thus $a = 1$, $b = 2007$, $c = 0$, and $d = 1$.

Now evaluate:

Therefore, the correct option is C.

Given: $Q = PAP^T$ and $P$ is an orthogonal matrix.

Find: $2a + b - 3c - 4d$ from $P^TQ^{2007}P$.

Instead of calculating $Q$ explicitly, use the fact that conjugation preserves powers:

So the whole problem reduces to finding $A^{2007}$.

For

each multiplication increases the upper-right entry by $1$. Hence,

Thus $a=1$, $b=2007$, $c=0$, $d=1$, and

Therefore, the answer is $2005$.