Question

$\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$ = $(a+b+c)^3$

Answer 1

$(a+b+c)^3$

Answer 2

To prove the identity:

$\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^3$

Let the given determinant be $D$.

$D = \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Apply the row operation $R_1 \to R_1 + R_2 + R_3$:

$D = \begin{vmatrix} (a-b-c)+2b+2c & 2a+(b-c-a)+2c & 2a+2b+(c-a-b) \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Simplify the elements of the first row:

$D = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Take out the common factor $(a+b+c)$ from $R_1$:

$D = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Apply the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$D = (a+b+c) \begin{vmatrix} 1 & 1-1 & 1-1 \\ 2b & (b-c-a)-2b & 2b-2b \\ 2c & 2c-2c & (c-a-b)-2c \end{vmatrix}$

Simplify the elements:

$D = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -a-b-c & 0 \\ 2c & 0 & -a-b-c \end{vmatrix}$

$D = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix}$

This is a lower triangular matrix. The determinant of a triangular matrix is the product of its diagonal elements.

$D = (a+b+c) \times [1 \times (-(a+b+c)) \times (-(a+b+c))]$

$D = (a+b+c) \times (a+b+c)^2$

$D = (a+b+c)^3$

Thus, the identity is proven.