Question

Let $P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 10 & 100 & -1 \end{pmatrix}$. Then $P^{4052}$ is equal to:

• A. $P$
• B. $P^T$
• C. $I$, the unit matrix of order 3
• D. $-P^T$
• E. $2P^T$

Hint:

If $A^2 = I$, then even powers give identity and odd powers give $A$.

Answer 1

The Correct Option is C

Concept:
• If $P^2 = I$, then $P^{\text{even}} = I$ and $P^{\text{odd}} = P$

Step 1: Compute $P^2$
\[ P^2 = P \cdot P = \begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 10 & 100 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 10 & 100 & -1 \end{pmatrix} \] Multiplying, \[ = \begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 10+(-10) & 100+(-100) & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{pmatrix} = I \]

Step 2: Use power property
\[ P^2 = I \Rightarrow P^{4052} = (P^2)^{2026} = I^{2026} = I \] Final Conclusion:
\[ P^{4052} = I \]