Prove that the determinant of the matrix [[b+c, a, a], [b, c+a, b], [c, c, a+b]] is 2abc.

Step 1: Start with the left-hand side (LHS) of the equation. Let the given determinant be . = b+c & a & a \\ b & c+a & b \\ c & c & a+b Step 2: Apply the row operation R_1 R_1 - R_2 - R_3. This operation changes the first row while keeping the determinant value unchanged. The new elements of the first row will be: • First element: (b+c) - b - c = 0 • Second element: a - (c+a) - c = a - c - a - c = -2c • Third element: a - b - (a+b) = a - b - a - b = -2b = 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b Step 3: Expand the determinant along the first row. The determinant of a 3 × 3 matrix a & b & c \\ d & e & f \\ g & h & i is a(ei-fh) - b(di-fg) + c(dh-eg). = 0 · c+a & b \\ c & a+b - (-2c) · b & b \\ c & a+b + (-2b) · b & c+a \\ c & c = 0 + 2c · (b(a+b) - b · c) - 2b · (b · c - c(c+a)) Step 4: Simplify the expression. = 2c · (ab + b^2 - bc) - 2b · (bc - c^2 - ac) = 2abc + 2b^2c - 2bc^2 - (2b^2c - 2bc^2 - 2abc) = 2abc + 2b^2c - 2bc^2 - 2b^2c + 2bc^2 + 2abc Step 5: Combine like terms. = (2abc + 2abc) + (2b^2c - 2b^2c) + (-2bc^2 + 2bc^2) = 4abc + 0 + 0 = 4abc This matches the right-hand side (RHS) of the given equation. Therefore, it is proven that: b+c & a & a \\ b & c+a & b \\ c & c & a+b = 4abc