Question

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

Hint:

To compute the determinant of a 3x3 matrix, use cofactor expansion and simplify the resulting expression.

Answer 1

Let \( A = \begin{pmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{pmatrix} \). We need to compute the determinant of matrix \( A \). The determinant is given by: \[ \text{det}(A) = (1 + a) \left| \begin{matrix} 1 + b & 1 \\ 1 & 1 + c \end{matrix} \right| - 1 \left| \begin{matrix} 1 & 1 \\ 1 & 1 + c \end{matrix} \right| + 1 \left| \begin{matrix} 1 & 1 + b \\ 1 & 1 \end{matrix} \right| \] Now, calculate each 2x2 determinant: 1. \[ \left| \begin{matrix} 1 + b & 1 \\ 1 & 1 + c \end{matrix} \right| = (1 + b)(1 + c) - 1 = 1 + b + c + bc - 1 = b + c + bc \] 2. \[ \left| \begin{matrix} 1 & 1 \\ 1 & 1 + c \end{matrix} \right| = (1)(1 + c) - (1)(1) = c \] 3. \[ \left| \begin{matrix} 1 & 1 + b \\ 1 & 1 \end{matrix} \right| = (1)(1) - (1)(1 + b) = -b \] Substitute these values into the determinant formula: \[ \text{det}(A) = (1 + a)(b + c + bc) - 1 \cdot c + 1 \cdot (-b) \] Simplify the expression: \[ \text{det}(A) = (1 + a)(b + c + bc) - c - b \] Expand: \[ \text{det}(A) = (1 + a)(b + c + bc) - (b + c) \] \[ \text{det}(A) = b + c + bc + ab + ac + abc - b - c \] Simplify: \[ \text{det}(A) = abc + ab + ac + bc \] Factor out \( abc \): \[ \text{det}(A) = abc(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \] Thus, we have proven the given identity.
Final Answer: \[ \boxed{\text{det}(A) = abc \left( 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)} \]