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Using Properties of Determinants, Prove That Matrix (Asquare2 + 2a,2a + 1,1,2a+1,A+2, 3, 3, 1 - Mathematics

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प्रश्न

Using properties of determinants, prove that 

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

योग
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उत्तर १

L.H.S = `|(a^2 + 2a,2a + 1,1),(2a+1,a+2, 1),(3, 3, 1)|`

`R_1 -> R_1 - R_2`

= `|(a^2-1, a-1,0),(2a+1,a+2,1),(3,3,1)|`

= `(a-1) |(a+1,1,0),(2a+1,a+2,1),(3,3,1)|`

`R_2 -> R_2 - R_2`

= `(a-1) |(a+1,1,0),(2a-2, a-1,0),(3,3,1)|`

= `(a-1)^2 |(a+1,1,0),(2,1,0),(3,3,1)|`

expanding along C3

= `(a-1)^2 (a+1-2) = (a-1)^2 (a-1)`

= `(a-1)^3 = R.H.S.`

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उत्तर २

`"L.H.S." = |(a^2+2a,2a+1,1),(2a+1,a+2,1),(3,3,1)|`

`R_2 ->R_2 - R_1, R_3-> R_3 -R_1`

= `|(a^2+2a,2a+1,1),(1-a^2,-a+1,0),(3-a^2-2a,3-2a-1,0)|`

= `|(a^2+2a,2a+1,1),(1-a^2,1-a,0),(3-a^2-2a, 2-2a,0)|`

Expanding along C3
= 1 [(1 - a2) (2 - 2a) - (1 - a) (3 - a2 - 2a)]
= 2 (1 -  a) (1 -  a) (1 + a) - (1 - a) (3 - a2 - 2a)
= (1 - a) [2 (1 - a2) - 3 + a2 + 2a]
= (1 - a) (2a - a2 - 1)
= (a - 1)3
= RHS

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2016-2017 (March) All India Set 1

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