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

May 1, 2017

I got: $- 8 \left(a + 1\right)$

#### Explanation:

Here we need to evaluate a Determinant.
Have a look:

May 1, 2017

Row reduction doesn't affect the determinant so here:

$\left(\begin{matrix}a & a + 1 & a + 2 \\ a + 1 & a + 3 & a + 1 \\ a + 2 & a + 1 & a\end{matrix}\right)$

R2 to R2 - R1; R3 to R3 - R1

$\left(\begin{matrix}a & a + 1 & a + 2 \\ 1 & 2 & - 1 \\ 2 & 0 & - 2\end{matrix}\right)$

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

$= 2 | \left(a + 1 , a + 2\right) , \left(2 , - 1\right) | - 0 | \text{whatever :)} | - 2 | \left(a , a + 1\right) , \left(1 , 2\right) |$

$= 2 \left(- \left(a + 1\right) - 2 \left(a + 2\right)\right) - 2 \left(2 a - \left(a + 1\right)\right)$

$= - 8 a - 8$

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

May 1, 2017

$- 8 \left(a + 1\right) .$

#### Explanation:

Let, $D = | \left(a , a + 1 , a + 2\right) , \left(a + 1 , a + 3 , a + 1\right) , \left(a + 2 , a + 1 , a\right) |$

We will simplify $D$ using the Elementary Operations.

Using ${C}_{2} - {C}_{1} \mathmr{and} {C}_{3} - {C}_{1} ,$ we get,

$D = | \left(a , 1 , 2\right) , \left(a + 1 , 2 , 0\right) , \left(a + 2 , - 1 , - 2\right) | ,$

$= 2 | \left(a , 1 , 1\right) , \left(a + 1 , 2 , 0\right) , \left(a + 2 , - 1 , - 1\right) | ,$

$= 2 | \left(a , 1 , 1\right) , \left(1 , 1 , - 1\right) , \left(2 , - 2 , - 2\right) | , \ldots \left[\because , {R}_{2} - {R}_{1} , {R}_{3} - {R}_{1}\right]$

$= 2 \left(2\right) | \left(a , 1 , 1\right) , \left(1 , 1 , - 1\right) , \left(1 , - 1 , - 1\right) | ,$

$= 4 | \left(a + 1 , 1 , 1\right) , \left(0 , 1 , - 1\right) , \left(0 , - 1 , - 1\right) | , \ldots \left[\because , {C}_{1} + {C}_{3}\right]$

$= 4 | \left(a + 1 , 0 , 0\right) , \left(0 , 0 , - 2\right) , \left(0 , - 1 , - 1\right) | , \ldots \left[\because , {R}_{1} + {R}_{3} , {R}_{2} + {R}_{3}\right]$

$= 4 \left(- 2\right) \left(- 1\right) | \left(a + 1 , 0 , 0\right) , \left(0 , 0 , 1\right) , \left(0 , 1 , 1\right) | ,$

$= 8 \left[\left(a + 1\right) \left(0 - 1\right) - 0 + 0\right] , \ldots \left[\because , \text{ expnsn. by } {R}_{1}\right]$

$\therefore D = - 8 \left(a + 1\right) .$

Enjoy Maths.!