Processing math: 100%


1)

If 1=[1a2a31b2b31c2c3]and2=[bcb+c1cac+a1aba+b1]

 then   12=


A) ab+bc+ca

B) abc

C) 2(ab+bc+ca)

D) (a+b+c)2

Answer:

Option A

Explanation:

1=[1a2a31b2b31c2c3]

R2R2R1,R3R3R1

=[1a2a30b2a2b3a30c2a2c3a3]

=(ba)(ca)[1a2a30b+ab2+a2+ab0c2+a2c2+a+ac]

=(ba)(ca)[(b+a)(c2+a2+ac)(c+a)(b2+a2+ab)]

=(ba)(ca)[bc2+a2b+abc+ac2+a3

+a2cb2ca2cabcb2aa3a2b]

=(ba)(ca)[(bc)(ab+bc+ca)]

=(ab)(bc)(ca)(ab+bc+ca)

 Now, 2=[bcb+c1cac+a1aba+b1]

 R2R2R1 and  R3R3R1

 =[bcb+c1c(ab)ab0b(ca)ac0]=(ab)(ca)[bcb+c1c10a10]

  =(ab)(ca)[1(cb)]

    =(ab)(cb)(ca)

Now, 12= (ab)(bc)(ca)(ab+bc+ca)(ab)(bc)(ca)

12 = (ab+bc+ca)