Question

Using properties of determinants, prove that 

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

Answer 1

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.`

Answer 2

`"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