Question

If \[ \left| \begin{matrix} x+a & y & z \\ x & y+b & z \\ x & y & z+c \end{matrix} \right| = 2abc, \] 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:

Whenever determinants contain repeated rows or columns with slight shifts, use row subtraction immediately to reveal hidden factorization patterns.

Answer 1

The Correct Option is B

Concept: Determinants having repeated variable structures are usually simplified using elementary row operations. Such problems test the ability to recognize hidden symmetry and reduce higher-order algebraic expressions efficiently.

Step 1: Writing the determinant carefully.
Given: \[ \Delta= \left| \begin{matrix} x+a & y & z \\ x & y+b & z \\ x & y & z+c \end{matrix} \right| \] and \[ \Delta=2abc \] We observe that each row contains repeated variables. Therefore row transformations will simplify the determinant significantly.

Step 2: Applying elementary row operations.
Apply: \[ R_1\to R_1-R_2 \] and \[ R_2\to R_2-R_3 \] Then the determinant becomes: \[ \Delta= \left| \begin{matrix} a & -b & 0 \\ 0 & b & -c \\ x & y & z+c \end{matrix} \right| \] Now the determinant becomes much easier to expand.

Step 3: Expanding along the first row.
Using cofactor expansion: \[ \Delta = a \left| \begin{matrix} b & -c \\ y & z+c \end{matrix} \right| -(-b) \left| \begin{matrix} 0 & -c \\ x & z+c \end{matrix} \right| \] Simplifying carefully: \[ \Delta = a[b(z+c)+cy] + b[cx] \] \[ = abz+abc+acy+bcx \] Thus, \[ \Delta = abc+abz+acy+bcx \] But given: \[ \Delta=2abc \] Hence, \[ abc+abz+acy+bcx = 2abc \] Subtracting \(abc\) from both sides: \[ abz+acy+bcx = abc \]

Step 4: Dividing throughout by \(abc\).
Since \(a,b,c\neq0\), divide throughout by \(abc\): \[ \frac{abz}{abc} + \frac{acy}{abc} + \frac{bcx}{abc} = \frac{abc}{abc} \] Thus, \[ \frac{z}{c} + \frac{y}{b} + \frac{x}{a} = 1 \] Rearranging: \[ \boxed{ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 } \]

Step 5: Alternative conceptual interpretation.
Observe that the determinant structure behaves like a shifted diagonal perturbation matrix. The extra determinant contribution beyond \(abc\) comes from the linear variable terms: \[ abz,\quad acy,\quad bcx \] Balancing this against the given determinant value immediately produces the normalized symmetric relation: \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \]