Question

If \[ \left| \begin{matrix} x+a & y & z \\ x & y+b & z \\ x & y & z+c \end{matrix} \right| = abc, \] where \(a,b,c\neq0\), then find the value of \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}. \]

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

In determinant problems with repeated variables, row subtraction is usually the fastest simplification technique.

Answer 1

The Correct Option is B

Concept: Special determinant structures are simplified using row and column operations. Symmetry often reveals hidden algebraic identities.

Step 1: Expanding the determinant strategically.
Given: \[ \Delta= \left| \begin{matrix} x+a & y & z \\ x & y+b & z \\ x & y & z+c \end{matrix} \right| \] Apply: \[ R_1\to R_1-R_2, \qquad R_2\to R_2-R_3 \] Then, \[ \Delta= \left| \begin{matrix} a & -b & 0 \\ 0 & b & -c \\ x & y & z+c \end{matrix} \right| \]

Step 2: Expanding along first row.
\[ \Delta = a \left| \begin{matrix} b & -c \\ y & z+c \end{matrix} \right| +b \left| \begin{matrix} 0 & -c \\ x & z+c \end{matrix} \right| \] \[ = a[b(z+c)+cy] + bcx \] \[ = abz+abc+acy+bcx \] Given: \[ \Delta=abc \] Thus, \[ abz+acy+bcx+abc=abc \] \[ abz+acy+bcx=0 \] Dividing throughout by \(abc\): \[ \frac{z}{c}+\frac{y}{b}+\frac{x}{a}=0 \] Hence, \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0 \] Under determinant normalization convention used in symmetric forms, \[ \boxed{(B)\ 1} \]