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Exercise 7.2 · Q14

Q.If A=[12α012]A = \begin{bmatrix} \dfrac{1}{2} & \alpha \\ 0 & \dfrac{1}{2} \end{bmatrix}, prove that ∑k=1ndet⁡(Ak)=13(1−14n).\displaystyle\sum_{k=1}^{n} \det(A^k) = \dfrac{1}{3}\left(1-\dfrac{1}{4^n}\right).

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det⁡(Ak)=(det⁡A)k=(1/4)k\det(A^k) = (\det A)^k = (1/4)^k; the sum is a geometric series with first term 1/41/4 and ratio 1/41/4.

We use the multiplicative property det⁡(Ak)=(det⁡A)k\det(A^k) = (\det A)^k and the geometric-series sum.

Step 1. AA is upper triangular, so det⁡A=12⋅12−α⋅0=14\det A = \dfrac12 \cdot \dfrac12 - \alpha\cdot 0 = \dfrac14.

Step 2. Therefore det⁡(Ak)=(det⁡A)k=(14)k\det(A^k) = (\det A)^k = \left(\dfrac14\right)^k.

Step 3. Sum the geometric series with first term 14\dfrac14 and common ratio 14\dfrac14: …

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