Question

Let A = `[(1, "a" ("b + c"), "bc"),(1, "b" ("c + a"), "ca"),(1, "c" ("a + b"), "ab")],` then Det. A is ____________.

पर्याय

• ab + bc + ca

• None of these

• 1 + ab + bc + ca

• 0

Answer 1

Let A = `[(1, "a" ("b + c"), "bc"),(1, "b" ("c + a"), "ca"),(1, "c" ("a + b"), "ab")],` then Det. A is 0.

Explanation:

`[(1, "a" ("b + c"), "bc"),(1, "b" ("c + a"), "ca"),(1, "c" ("a + b"), "ab")] => "det. A" = [(1, "a" ("b + c"), "bc"),(1, "b" ("c + a"), "ca"),(1, "c" ("a + b"), "ab")],`

Apply C₂ → C₂ + C₃

`=> "det.A" = abs((1, "ab + ac + bc", "bc"),(1, "bc + ca + ab", "ca"),(1, "ca + bc + ab",  "ab"))`

`=> ("ab + bc + ca") abs((1,1,"bc"),(1,1,"ca"),(1,1,"ab")) = 0`