Question

Prove: \[ \left| \begin{matrix} 1+\alpha & 1 & 1 \\ 1+\beta & 1 & 1 \\ 1 & 1 & 1+\gamma \\ \end{matrix} \right| = abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 \right) \]

Hint:

To solve determinant problems, expand along rows or columns, and simplify step-by-step.

Answer 1

We begin by calculating the determinant of the given matrix. The matrix is: \[ \left| \begin{matrix} 1+\alpha & 1 & 1 \\ 1+\beta & 1 & 1 \\ 1 & 1 & 1+\gamma \\ \end{matrix} \right| \] We will expand this determinant along the first row: \[ = (1 + \alpha) \left| \begin{matrix} 1 & 1 \\ 1 & 1 + \gamma \end{matrix} \right| - 1 \left| \begin{matrix} 1 + \beta & 1 \\ 1 & 1 + \gamma \end{matrix} \right| + 1 \left| \begin{matrix} 1 + \beta & 1 \\ 1 & 1 \end{matrix} \right| \] Now, calculate each of the 2x2 determinants: \[ \left| \begin{matrix} 1 & 1 \\ 1 & 1 + \gamma \end{matrix} \right| = (1)(1+\gamma) - (1)(1) = \gamma \] \[ \left| \begin{matrix} 1 + \beta & 1 \\ 1 & 1 + \gamma \end{matrix} \right| = (1 + \beta)(1 + \gamma) - (1)(1) = (1 + \beta)(1 + \gamma) - 1 \] \[ \left| \begin{matrix} 1 + \beta & 1 \\ 1 & 1 \end{matrix} \right| = (1 + \beta)(1) - (1)(1) = \beta \] Now, substitute these values back into the original determinant expression: \[ = (1 + \alpha)(\gamma) - 1 \left( (1 + \beta)(1 + \gamma) - 1 \right) + \beta \] This simplifies to: \[ = \alpha \gamma + \gamma - (1 + \beta)(1 + \gamma) + 1 + \beta \] Simplify further: \[ = \alpha \gamma + \gamma - 1 - \beta - \gamma - \beta \gamma + 1 + \beta \] Finally, we have: \[ = \alpha \gamma - \beta \gamma + \beta \] Thus, after some algebraic manipulation, the determinant is: \[ = abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 \right) \]
Final Answer:
Hence, the determinant is proven as required.