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shivusuja

Member

Registered: 2006-09-14

Posts: 56

Algebra

Please Help Me Out Of This Sum
Using Property Of  Determinant And Without Expanding Prove That
[1       1+p        1+p+q          ]
[2       3+2p      1+3p+2q      ]    =i
[3       6+3p       1+6p+3q     ]

Bob

Administrator

Registered: 2010-06-20

Posts: 10,164

Re: Algebra

hi shivusuja

You need to do row and column operations such as

{new column 2} = {old column 2} - {old column 1}

That will make the second element just 'p' then

{new row 2} = {old row 2} - 2 x {old row 1}

Manipulations like this will allow you to eliminate the ps and qs and put lots of zeros into element spaces until you have left just a non zero leading diagonal.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

AASTHA DUA

Member

Registered: 2012-07-09

Posts: 1

Re: Algebra

its answer is one i.e. 1
HERE IS THE SOL. C IS FOR COLUMN
1. C 3 = C 3 - q. C 1
2. C 2 = C 2 - p. C 1
3. C 3 = C 3 - P. C 2

enjoy

Member

Registered: 2012-07-18

Posts: 1

Re: Algebra

good solution