Question

`abs ((1 + "a", "b", "c"),("a", 1 + "b", "c"),("a", "b", 1 + "c")) =` ____________

पर्याय

• abc

• None of these

• a + b + c

• 1 + a + b + c

Answer 1

`abs ((1 + "a", "b", "c"),("a", 1 + "b", "c"),("a", "b", 1 + "c")) =` 1 + a + b + c

Explanation:

`abs ((1 + "a", "b", "c"),("a", 1 + "b", "c"),("a", "b", 1 + "c"))`

Apply, C₁→ C₁ + C₂ + C₃

`=> abs ((1 + "a + b + c", "b" , "c"),(1 + "a + b + c", 1 + "b", "c"),(1 + "a + b + c", "b", 1 + "c"))`

`=> (1 + "a + b + c") abs ((1, "b", "c"),(1,1 + "b", c),(1,"b", 1 + "c"))`

Apply R₁→R₁ - R₂

`=> (1 + "a + b + c") abs ((0, -1, 0),(1,1 + "b", "c"),(1,"b", 1 + "c"))`

now expanding along R₁

`=>` (1 + a + b + c)(1(1 + c - c))

= 1 + a + b + c