Solving a determinant to give it's factors

[nirajnishad]

Homework Statement

I have a determinant to solve.

Fistrow is -2a, a + b, a+c
Second row is b + a, -2b, b + c
Third row is c+a, c+b, -2c

Prove that the determinant is equal to 4 (b +c) (c +a) (a + b)

Homework Equations

Not applicable.

The Attempt at a Solution

I have tried addig up rows and column both three at a time and two at a time. I am unable to find any common factor. This is one of my first sums, so I'msure I'm missing out on something.


Also, could you tell me how to enter a determinant at PhysicsForums?

 

[scottie_000]

Fistrow is -2a, a + b, a+c
Second row is b + a, -2b, b + c
Third row is c+a, c+b, -2c

The best way to write this is using $$\LaTeX$$, but that's best learned another day.

What operations for finding a determinant do you know about so far?

Hint: You want to try eliminating entries in one of the rows so they become 0. What about replacing the 3rd row with (3rd+2nd+1st)?

 

[hunt_mat]

So you have your determinant:
$$\left|\begin{array}{ccc}<br /> -2a & a+b & a+c \\<br /> b+a & -2b & b+c \\<br /> c+a & c+b & -2c<br /> \end{array}\right|$$
Why not just expant the determinant? It's small enough to do by hand and have in mind to keep in one of the factors in the solution. It's a cheat but it will give you some idea of determinants.

 

[nirajnishad]

The best way to write this is using $$\LaTeX$$, but that's best learned another day.

What operations for finding a determinant do you know about so far?

Hint: You want to try eliminating entries in one of the rows so they become 0. What about replacing the 3rd row with (3rd+2nd+1st)?

i have tried adding all rows and colums.
after doing it ,i am unable to find any common factor in any row or coloumn

 

[Mark44]

You can get fancy and perform row operations to make the determinant simpler, or you can just use brute force and expand by cofactors.

For example, expanding across the top row:
$$\left|\begin{array}{ccc} 1 & 2 & 0 \\3 & -2 & 1 \\2 & 0 & -2\end{array}\right|= 1\left|\begin{array}{cc} -2 & 1 \\0 & -2 \end{array}\right| - 2 \left|\begin{array}{cc} 3 & 1 \\2 & -2 \end{array}\right| + 0\left|\begin{array}{cc} 3 & -2 \\2 & 0 \end{array}\right|$$

Now it's a matter of evaluating three 2 x 2 determinants.