Let A = [3 -8; 7 k]. For A to have 0 as an eigenvalue, k must be
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det(A) = (3-k)(-8) = -8k + 24 For A to have 0 as an eigenvalue, det(A) must be equal to 0. -8k + 24 = 0 Solving for k: -8k = -24 k = 3 Show more…
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