IIT JEE , AIEEE Forum - Mathematics , Algebra

vaibhavvishen

Q: Using the properties of determinants prove this:

1+a²-b² 2ab -2b

2ab 1-a²+b² 2a = (1+a²+b²)³

2b -2a 1-a²-b²

lazycol

1+a²-b² 2ab -2b =(1+a²+b²) * 1 2ab -2b

2ab 1-a²+b² 2a 0 1-a²+b² 2a

2b -2a 1-a²-b² b -2a 1-a²-b²

[ C1 = C1 - bC3 ]

=(1+a²+b²)²* 1 0 -2b = (1+a²+b²) * 1 0 -2b

0 1 2a 0 1 2a

b -a 1-a²-b² 0 0 1

[ C2 = C2 + aC3 ] [R3 = R3 -bR1 - aR2]

=(1+a²+b²)³

puneet

hey

this is a simple one ..

replace C1 by C1 - b(C3) and C2 by C2 + a(C3)

Now take out (1+a²+b²) common from C1 and C2.

The determinant has now reduced considerably and it can now be easily solved.

I hope this helps

cheers

vinkymarali

Hi Vaibhav,

I'm telling u a short method of checking such sums' L.H.S. & R.H.S. r equal ...

Put a, b as any number preferably 0
then solve the determinant ..

| 1 0 0 |
| 0 1 0 | = (1)^3
| 0 0 0 |

L.H.S = 1(1-0)
=1
R.H.S =1

So, L.H.S = R.H.S ...

After practising a few sums u'll b able 2 solve such sums orally ..

This method helps in exams which has objective type pattern ..

Hope u liked the shortcut 2 solve such problems ...

Pleazz Rate me .... :)

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