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Let A be the matrix [5, 5α, α; 0, α, 5α; 0, 0, 5]. If |A²| = 25, then the value of |α| is

1. (1)/(5)
2. 5
3. 5²
4. 1

Correct answer: (1)/(5)

Solution

A is upper triangular with diagonal 5, alpha, 5, so det(A) = 25*alpha. Then |A^2| = (det A)^2 = 625*alpha^2 = 25 -> alpha^2 = 1/25 -> |alpha| = 1/5.

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