Question

The determinant `abs (("a","bc","a"("b + c")),("b","ac","b"("c + a")),("c","ab","c"("a + b"))) =` ____________

पर्याय

• None of these

• (ab + bc + ca ) (a – b) (b – c) (c – a)

• abc (a – b) (b – c) (c – a)

• abc (ab + bc + ca)

Answer 1

The determinant `abs (("a","bc","a"("b + c")),("b","ac","b"("c + a")),("c","ab","c"("a + b"))) =` (ab + bc + ca ) (a – b) (b – c) (c – a)

Explanation:

`abs (("a","bc","a"("b + c")),("b","ac","b"("c + a")),("c","ab","c"("a + b"))) => abs (("a","bc","ab" + "ac"),("b","ac","bc" + "ab"),("c","ab","ca" + "bc"))`

Apply, C₃→C₃ + C₂,

`abs (("a","bc","ab" + "bc" + "ac"),("b","ac","bc" + "ab" + "ca"),("c","ab","ca" + "bc" + "ab"))`

`=> ("ab" + "bc" + "ca") abs (("a", "bc",1),("b", "ac", 1),("c", "ab", 1))`

Apply, R₁ → R₁ - R₂, R₂ → R₂ - R₃

`("ab" + "bc" + "ca") abs (("a - b", "-c" ("a - b"), 0),("b - c", "-a" ("b - c"), 0),("c", "ab", 1))`

`("a - b"),("b - c") ("ab + bc + ca") abs ((1,"-c",0),(1,"-a",0),("c","ab",1))`

expanding along C₃

`=>("a - b")("b - c") ("ab + bc + ca") ("-a + c")`

`=>("a - b")("b - c")("c - a")("ab + bc + ca")`