Ex 4.2, 5

Transcript

Ex 4.2, 5 Using the property of determinants and without expanding, prove that: b+c﷮q+r﷮y+z﷮c+a﷮r+p﷮z+x﷮a+b﷮p+q﷮x+y﷯﷯ = 2 a﷮p﷮x﷮b﷮q﷮y﷮c﷮r﷮z﷯﷯ Taking L.H.S b+c﷮q+r﷮y+z﷮c+a﷮r+p﷮z+x﷮a+b﷮p+q﷮x+y﷯﷯ Applying R3 → R3 + R2 + R1 = b+𝑐﷮𝑞+𝑟﷮𝑦+𝑧﷮c+a﷮r+𝑝﷮z+x﷮a+b+c+a+b+c﷮𝑝+𝑞+𝑟+𝑞+𝑞+𝑟﷮x+𝑦+𝑧+𝑥+𝑦+𝑧﷯﷯ = b+𝑐﷮𝑞+𝑟﷮𝑦+𝑧﷮c+a﷮r+𝑝﷮z+x﷮𝟐(a+b+c)﷮𝟐(𝑝+𝑞+𝑟)﷮𝟐(x+𝑦+𝑧)﷯﷯ Taking 2 common from R3 = 2 b+𝑐﷮𝑞+𝑟﷮𝑦+𝑧﷮c+a﷮r+𝑝﷮z+x﷮(a+b+c)﷮(𝑝+𝑞+𝑟)﷮(x+𝑦+𝑧)﷯﷯ Applying R1 → R1 – R3 = 2 b+𝑐−𝑎−𝑏−𝑐﷮𝑞+𝑟−𝑝−𝑞−𝑟﷮𝑦+𝑧−𝑥−𝑦−𝑧﷮c+a﷮r+𝑝﷮z+x﷮a+b+c﷮𝑝+𝑞+𝑟﷮x+𝑦+𝑧﷯﷯ = 2 −𝑎﷮−𝑝﷮−𝑥﷮c+a﷮r+𝑝﷮z+x﷮a+b+c﷮𝑝+𝑞+𝑟﷮x+𝑦+𝑧﷯﷯ Applying R2 → R2 – R3 = 2 −𝑎﷮−𝑝﷮−𝑥﷮c+a−a−b−c﷮r+𝑝−𝑝−𝑞−𝑟﷮z+x−x−y−z﷮a+b+c﷮𝑝+𝑞+𝑟﷮x+𝑦+𝑧﷯﷯ = 2 −𝑎﷮−𝑝﷮−𝑥﷮−b﷮−𝑞﷮−y﷮a+b+c﷮𝑝+𝑞+𝑟﷮x+𝑦+𝑧﷯﷯ Applying R3 → R3 + R1 + R2 = 2 −𝑎﷮−𝑝﷮−𝑥﷮−b﷮−𝑞﷮−y﷮a+b+c−𝑎−𝑏﷮𝑝+𝑞+𝑟−𝑝−𝑞﷮x+𝑦+𝑧−𝑥−𝑦﷯﷯ = 2 −𝑎﷮−𝑝﷮−𝑥﷮−b﷮−𝑞﷮−y﷮𝑐﷮𝑟﷮𝑧﷯﷯ Taking –1 Common from R1 & R3 = 2 ( –1) ( –1) 𝑎﷮𝑝﷮𝑥﷮b﷮𝑞﷮y﷮𝑐﷮𝑟﷮𝑧﷯﷯ = 2 𝑎﷮𝑝﷮𝑥﷮b﷮𝑞﷮y﷮𝑐﷮𝑟﷮𝑧﷯﷯ = R.H.S Hence Proved