The value of \(\begin{vmatrix}2a + b + c & b & c\\ a & a + 2b + c & c\\ a & b & a + b + 2c\end{vmatrix}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2(a + b + c)^{3}\)
In every row the three elements add up to \(2(a + b + c)\). Apply \(C_1 \to C_1 + C_2 + C_3\) and take this common factor out of the first column:
\(\displaystyle \Delta = 2(a + b + c)\begin{vmatrix}1 & b & c\\ 1 & a + 2b + c & c\\ 1 & b & a + b + 2c\end{vmatrix}\)
Apply \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\(\displaystyle \Delta = 2(a + b + c)\begin{vmatrix}1 & b & c\\ 0 & a + b + c & 0\\ 0 & 0 & a + b + c\end{vmatrix}\)
The new determinant is upper triangular, so it is the product of its diagonal elements:
\(\Delta = 2(a + b + c) \cdot (a + b + c)^{2}\)
\(\Delta = 2(a + b + c)^{3}\)
Check with a = b = c = 1: the determinant is \(\begin{vmatrix}4 & 1 & 1\\ 1 & 4 & 1\\ 1 & 1 & 4\end{vmatrix} = 54 = 2 \times 3^{3}\).