GATE 2024 IN Q.26: solution and answer

Official answer: -2Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. The determinant of a square matrix equals the product of its eigenvalues (counted with multiplicity). A $3\times 3$ matrix has exactly three eigenvalues, and here they are $\frac{1}{4}$, $1$ and $-2$, so $$|P|={\lambda }_1{\lambda }_2{\lambda }_3=\frac{1}{4}\times 1\times (-2)=-\frac{1}{2}.$$

2. No eigenvalue is zero, so $|P|\ne 0$ and $P$ is invertible.

3. If $P\mathbf{x}=\lambda \mathbf{x}$ with $\lambda \ne 0$, then $\mathbf{x}=\lambda P^{-1}\mathbf{x}$, so $P^{-1}\mathbf{x}=\frac{1}{\lambda }\mathbf{x}$. The eigenvalues of $P^{-1}$ are therefore the reciprocals: $4,1,-\frac{1}{2}$.

4. Hence $$|P^{-1}|=4\times 1\times \left(-\frac{1}{2}\right)=-2.$$ The same result follows from $|P^{-1}|=\frac{1}{|P|}=\frac{1}{-1/2}=-2$.

5. Here $|P^{-1}|$ is determinant notation (not absolute value), so the value is $-2$. It is already an integer, so rounding changes nothing.