Question

question 8
let a be a 2×2 matrix so that ( p^{-1}ap = d ), where ( d = \begin{bmatrix} 1 & 0 \\ 0 & -3 end{bmatrix} ) and ( p = \begin{bmatrix} 1 & -1 \\ 1 & 1 end{bmatrix} ).
which of the following is ( a^{2025} )?
(a) ( a^{2025} = \frac{1}{2}\begin{bmatrix} -1 + 3^{2025} & 1 + 3^{2025} \\ 1 + 3^{2025} & -1 + 3^{2025} end{bmatrix} ).
(b) ( a^{2025} = \frac{1}{2}\begin{bmatrix} 1 - 3^{2025} & 1 + 3^{2025} \\ 1 + 3^{2025} & 1 - 3^{2025} end{bmatrix} ).
(c) ( a^{2025} = \frac{1}{2}\begin{bmatrix} 1 + 3^{2025} & 1 - 3^{2025} \\ 1 - 3^{2025} & 1 + 3^{2025} end{bmatrix} ).
(d) ( a^{2025} = \frac{1}{2}\begin{bmatrix} -1 - 3^{2025} & 1 + 3^{2025} \\ 1 + 3^{2025} & -1 - 3^{2025} end{bmatrix} ).
(e) none of the above.

Explanation:

Step1: Recall Matrix Diagonalization Property

If \( P^{-1}AP = D \), then \( A = PDP^{-1} \), and \( A^n = PD^nP^{-1} \) for any positive integer \( n \). First, we need to find \( P^{-1} \).

For a \( 2\times2 \) matrix \( P=

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\), the inverse is \( P^{-1}=\frac{1}{\det(P)}

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

\), where \( \det(P)=ad - bc \).

For \( P=

$$\begin{bmatrix}1&-1\\1&1\end{bmatrix}$$

\), \( \det(P)=(1)(1)-(-1)(1)=1 + 1 = 2 \). So, \( P^{-1}=\frac{1}{2}

$$\begin{bmatrix}1&1\\-1&1\end{bmatrix}$$

\).

Step2: Find \( D^{2025} \)

Given \( D=

$$\begin{bmatrix}1&0\\0&-3\end{bmatrix}$$

\), for a diagonal matrix, \( D^n=

$$\begin{bmatrix}1^n&0\\0&(-3)^n\end{bmatrix}$$

\). So, \( D^{2025}=

$$\begin{bmatrix}1^{2025}&0\\0&(-3)^{2025}\end{bmatrix}$$

=

$$\begin{bmatrix}1&0\\0&-3^{2025}\end{bmatrix}$$

\) (since \( 2025 \) is odd, \( (-3)^{2025}=-3^{2025} \)).

Step3: Compute \( A^{2025}=PD^{2025}P^{-1} \)

First, compute \( PD^{2025} \):
\( PD^{2025}=

$$\begin{bmatrix}1&-1\\1&1\end{bmatrix}$$

$$\begin{bmatrix}1&0\\0&-3^{2025}\end{bmatrix}$$

=

$$\begin{bmatrix}1\times1+(-1)\times0&1\times0+(-1)\times(-3^{2025})\\1\times1+1\times0&1\times0+1\times(-3^{2025})\end{bmatrix}$$

=

$$\begin{bmatrix}1&3^{2025}\\1&-3^{2025}\end{bmatrix}$$

\)

Then, multiply by \( P^{-1}=\frac{1}{2}

$$\begin{bmatrix}1&1\\-1&1\end{bmatrix}$$

\):
\( A^{2025}=

$$\begin{bmatrix}1&3^{2025}\\1&-3^{2025}\end{bmatrix}$$

\times\frac{1}{2}

$$\begin{bmatrix}1&1\\-1&1\end{bmatrix}$$

\)
\( =\frac{1}{2}

$$\begin{bmatrix}1\times1 + 3^{2025}\times(-1)&1\times1 + 3^{2025}\times1\\1\times1+(-3^{2025})\times(-1)&1\times1+(-3^{2025})\times1\end{bmatrix}$$

\)
\( =\frac{1}{2}

$$\begin{bmatrix}1 - 3^{2025}&1 + 3^{2025}\\1 + 3^{2025}&1 - 3^{2025}\end{bmatrix}$$

\)

Answer:

B. \( A^{2025}=\frac{1}{2}

$$\begin{bmatrix}1 - 3^{2025}&1 + 3^{2025}\\1 + 3^{2025}&1 - 3^{2025}\end{bmatrix}$$

\)