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")|`

Answer 1

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.