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

3 Answers
Gió
May 1, 2017

I got: -8(a+1)

Explanation:

Here we need to evaluate a Determinant.
Have a look:
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Eddie
May 1, 2017

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.

Ratnaker Mehta
May 1, 2017

-8(a+1).

Explanation:

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.!

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