Question

Prove that: \[ \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = abc \left( 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \]

Hint:

Expand determinants systematically and use minor expansions to simplify step-by-step.

Answer 1

Expand the determinant along the first row: \[ \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = (1+a) \begin{vmatrix} 1+b & 1 \\ 1 & 1+c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1+c \end{vmatrix} + 1 \begin{vmatrix} 1 & 1+b \\ 1 & 1 \end{vmatrix}. \] Calculate each minor: \[ \begin{vmatrix} 1+b & 1 \\ 1 & 1+c \end{vmatrix} = (1+b)(1+c) - 1 = bc + b + c, \] \[ \begin{vmatrix} 1 & 1 \\ 1 & 1+c \end{vmatrix} = c, \] \[ \begin{vmatrix} 1 & 1+b \\ 1 & 1 \end{vmatrix} = -b. \] Substitute back: \[ \text{Determinant} = (1+a)(bc+b+c) - c - b. \] Simplify and verify: \[ \text{Determinant} = abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \]