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Question
Choose the correct alternative:
if Δ = `|("a", "b", "c"),(x, y, z),("p", "q", "r")|` then `|("ka", "kb","kc"),("k"x, "k"y, "k"z),("kp", "kq", "kr")|` is
Options
Δ
kΔ
3kΔ
k3Δ
Solution
k3Δ
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