Questions: Let A=PDP^-1 and P and D as shown below. Compute A^4. P=[ 1 4 2 7 ], D=[ 2 0 0 3 ] A^4= (Simplify your answer.)

Transcript text: Let $A=P D P^{-1}$ and $P$ and $D$ as shown below. Compute $A^{4}$. \[ P=\left[\begin{array}{ll} 1 & 4 \\ 2 & 7 \end{array}\right], D=\left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right] \] \[ A^{4}= \] $\square$ (Simplify your answer.)

Solution

Solution Steps

Step 1: Define Matrices

Let $ P = \begin{array}{cc} 1 & 4 \\ 2 & 7 \end{array} $ and $ D = \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} $.

Step 2: Compute $ D^4 $

Calculate $ D^4 $: \[ D^4 = \begin{array}{cc} 2^4 & 0 \\ 0 & 3^4 \end{array} = \begin{array}{cc} 16 & 0 \\ 0 & 81 \end{array} \]

Step 3: Compute $ P^{-1} $

Find the inverse of $ P $: \[ P^{-1} = \frac{1}{\text{det}(P)} \text{adj}(P) \] where $ \text{det}(P) = 1 \cdot 7 - 4 \cdot 2 = -1 $ and $ \text{adj}(P) = \begin{array}{cc} 7 & -4 \\ -2 & 1 \end{array} $. Thus, \[ P^{-1} = \begin{array}{cc} -7 & 4 \\ 2 & -1 \end{array} \]

Step 4: Compute $ A^4 $

Now compute $ A^4 = P D^4 P^{-1} $: \[ A^4 = \begin{array}{cc} 1 & 4 \\ 2 & 7 \end{array} \begin{array}{cc} 16 & 0 \\ 0 & 81 \end{array} \begin{array}{cc} -7 & 4 \\ 2 & -1 \end{array} \]

Step 5: Final Calculation

Perform the matrix multiplication:

First, compute $ P D^4 $: \[ P D^4 = \begin{array}{cc} 1 \cdot 16 + 4 \cdot 0 & 1 \cdot 0 + 4 \cdot 81 \\ 2 \cdot 16 + 7 \cdot 0 & 2 \cdot 0 + 7 \cdot 81 \end{array} = \begin{array}{cc} 16 & 324 \\ 32 & 567 \end{array} \]
Then, compute $ (P D^4) P^{-1} $: \[ A^4 = \begin{array}{cc} 16 & 324 \\ 32 & 567 \end{array} \begin{array}{cc} -7 & 4 \\ 2 & -1 \end{array} = \begin{array}{cc} 16 \cdot -7 + 324 \cdot 2 & 16 \cdot 4 + 324 \cdot -1 \\ 32 \cdot -7 + 567 \cdot 2 & 32 \cdot 4 + 567 \cdot -1 \end{array} \] Calculating each entry gives: \[ A^4 = \begin{array}{cc} -112 + 648 & 64 - 324 \\ -224 + 1134 & 128 - 567 \end{array} = \begin{array}{cc} 536 & -260 \\ 910 & -439 \end{array} \]