Topic: Engineering Mathematics
If $A=\begin{pmatrix}9&6\\ 8&7\end{pmatrix}$ then det $(A^{99}-A^{98})$ is
- A. 1
- B. 48
- C. 0
- D. $2^{99}$
Answer: C. 0
Explanation: We know that $|AB|=|A||B|$. Thus, $|A^{99}-A^{98}| = |A^{98}(A-I)| = |A^{98}||A-I|$. Let's evaluate $|A-I| = \begin{vmatrix} 9-1 & 6 \\ 8 & 7-1 \end{vmatrix} = \begin{vmatrix} 8 & 6 \\ 8 & 6 \end{vmatrix} = (8 \times 6) - (6 \times 8) = 48 - 48 = 0$. Since $|A-I|=0$, $|A^{99}-A^{98}| = |A^{98}| \times 0 = 0$.