Question

Prove that \[ \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3 \]

Hint:

Column/row transformations often simplify determinants to triangular form.

Answer 1

Step 1: Expand determinant.
Let \[ \Delta = \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} \] Step 2: Apply column transformation.
Perform $C_1 \to C_1 + C_2 + C_3$: \[ \Delta = \begin{vmatrix} (a+b+2c)+a+b & a & b \\ c+(b+c+2a)+b & b+c+2a & b \\ c+a+(c+a+2b) & a & c+a+2b \end{vmatrix} \] \[ \Delta = \begin{vmatrix} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2a & b \\ 2(a+b+c) & a & c+a+2b \end{vmatrix} \] Step 3: Factor common term.
Take $2(a+b+c)$ common from $C_1$: \[ \Delta = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix} \] Step 4: Expand determinant.
Apply $R_2 \to R_2 - R_1$, $R_3 \to R_3 - R_1$: \[ = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix} \] Step 5: Simplify.
\[ = 2(a+b+c)\big[(b+c+a)(c+a+b)\big] \] \[ = 2(a+b+c)(a+b+c)(a+b+c) \] \[ = 2(a+b+c)^3 \]

Final Answer: \[ \boxed{\Delta = 2(a+b+c)^3} \]