How you calculate this? #abs((a,a+1,a+2),(a+1,a+3,a+1),(a+2,a+1,a))#

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Answer 1 · 2017-05-01T13:20:45

I got: #-8(a+1)#

Here we need to evaluate a Determinant.
Have a look:
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Answer 2 · 2017-05-01T14:13:23

Row reduction doesn't affect the determinant so here:

#((a,a+1,a+2),(a+1,a+3,a+1),(a+2,a+1,a))#

#R2 to R2 - R1; R3 to R3 - R1#

#((a,a+1,a+2),(1,2,-1),( 2,0,-2))#

Now take the determinant from the bottom row as it's got a zero:

#= 2 |(a+1, a+2),(2, -1)| - 0 |"whatever :)"| - 2|(a, a+1),(1, 2)|#

#= 2 (-(a+1) - 2 (a+2)) - 2(2 a - (a+1))#

#= -8a - 8 #

This matrix is symmetric but that doesn't help when it comes to determinants.

Answer 3 · 2017-05-01T14:40:48

#-8(a+1).#

Let, #D=|(a,a+1,a+2),(a+1,a+3,a+1),(a+2,a+1,a)|#

We will simplify #D# using the Elementary Operations.

Using #C_2-C_1 and C_3-C_1,# we get,

#D=|(a,1,2),(a+1,2,0),(a+2,-1,-2)|,#

#=2|(a,1,1),(a+1,2,0),(a+2,-1,-1)|,#

#=2|(a,1,1),(1,1,-1),(2,-2,-2)|,...[because, R_2-R_1,R_3-R_1]#

#=2(2)|(a,1,1),(1,1,-1),(1,-1,-1)|,#

#=4|(a+1,1,1),(0,1,-1),(0,-1,-1)|,...[because, C_1+C_3]#

#=4|(a+1,0,0),(0,0,-2),(0,-1,-1)|,...[because, R_1+R_3, R_2+R_3]#

#=4(-2)(-1)|(a+1,0,0),(0,0,1),(0,1,1)|,#

#=8[(a+1)(0-1)-0+0],...[because," expnsn. by "R_1]#

#:. D=-8(a+1).#

Enjoy Maths.!