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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`
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.`
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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