Question

Use the factorization A=PDP^{-1} to compute A^{k}.

Polynomial factorization
ANSWERED
asked 2021-05-27

Use the factorization \(A=PDP^{-1}\) to compute \(A^{k}\). where k represents an arbitrary integer.
\(\begin{bmatrix}a & 11(b-a) \\0 & b \end{bmatrix}=\begin{bmatrix}1 & 11 \\0 & 1 \end{bmatrix}\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}\)
\(A^{k}=\)?

Answers (1)

2021-05-28
Step 1
Given
\(\begin{bmatrix}a & 11(b-a) \\0 & b \end{bmatrix}=\begin{bmatrix}1 & 11 \\0 & 1 \end{bmatrix}\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}\)
To find
\(A^{k}=\)?
Step 2
We know that,
\(A^{k}=PD^{k}P^{-1}=\begin{bmatrix}1 & 11 \\0 & 1 \end{bmatrix}\begin{bmatrix}a^{k} & 0 \\0 & b^{k} \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}\)
\(=\begin{bmatrix}a^{k} & 11b^{k} \\0 & b^{k} \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}\)
\(=\begin{bmatrix}a^{k} & -11a^{k}+11b^{k} \\0 & b^{k} \end{bmatrix}\)
\(=\begin{bmatrix}a^{k} & 11b^{k}-11a^{k} \\0 & b^{k} \end{bmatrix}\)
Hence,
\(A^{k}=\begin{bmatrix}a^{k} & 11(b^{k}-a^{k}) \\0 & b^{k} \end{bmatrix}\)
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