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# Using Properties of Determinants, Prove That: |(1+A^2-b^2, 2ab, -2b),(2ab, 1-a^2+B^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + A^2 + B^2)^3 - ISC (Science) Class 12 - Mathematics

ConceptProperties of Determinants

#### Question

Using properties of determinants, prove that:

|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + a^2 + b^2)^3

#### Solution

|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + a^2 + b^2)^3

Let

Delta = |(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)|

We shall try to introduce zeros at a many places as possible keeping in mind that we have to introduce the factor 1 + a^2 +  b^2

Applying C_1 -> C_1 - bC_3 and C_2 -> C_2 +  aC_3 we get

Delta = |(1+a^2+b^2, 0, -2b),(0, 1+a^2+b^2, 2a), (b(1+a^2 +b^2), -a(1+a^2+b^2), 1-a^2 - b^2)|

=> Delta = (1 + a^2  +b^2)^2 |(1,0,-2b),(0,1,2a),(b, -a, -a^2-b^2)|   [Taking (1+ a^2 + b^2) common from both C_1 and C_2]

=> Delta = (1+ a^2 + b^2)^2 |(1,0,-2b),(0,1, 2a),(0,0, 1+z^2+b^2) |       [Applying R_3 -> R_3- bR_1 + aR_2]

=> Delta= (1 + a^2 + b^2)^2 xx 1 xx |(1, 2a),(0, 1+a^2 + b^2)|       [Expanding along C_1]

=> Delta= (1 + a^2 + b^2 )^3

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#### APPEARS IN

2014-2015 (March) (with solutions)
Question 2.1 | 5.00 marks
Solution Using Properties of Determinants, Prove That: |(1+A^2-b^2, 2ab, -2b),(2ab, 1-a^2+B^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + A^2 + B^2)^3 Concept: Properties of Determinants.
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