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Question
Answer the following question:
Without expanding determinant show that
`|(l, "m", "n"),("e", "d", "f"),("u", "v", "w")| = |("n", "f", "w"),(l, "e", "u"),("m", "d", "v")|`
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Solution
L.H.S. = `|(l, "m", "n"),("e", "d", "f"),("u", "v", "w")|`
Interchanging rows and columns, we get
L.H.S. = `|(l, "e", "u"),("m", "d", "v"),("n", "f", "w")|`
Applying R2 ↔ R3, we get
L.H.S. = `-|(l, "e", "u"),("n", "f", "w"),("m", "d", "v")|`
Applying R1 ↔ R2, we get
L.H.S. = `|("n", "f", "w"),(l, "e", "u"),("m", "d", "v")|`
= R.H.S.
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