Question

If \(x,y,z\) are different from zero and
\[ \Delta= \begin{vmatrix} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{vmatrix}=0 \] then the value of the expression \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\) is

• A. 0
• B. -1
• C. 1
• D. 2

Hint:

In determinants, use row operations to create zeros and simplify quickly before expanding.

Answer 1

The Correct Option is D

Step 1: Observe the structure of determinant.
Given determinant is:
\[ \Delta= \begin{vmatrix} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{vmatrix} \]
Step 2: Expand determinant by applying row/column operations.
Take differences to simplify:
Subtract \(R_2\) from \(R_3\):
\[ R_3 \to R_3-R_2 \Rightarrow (0,-y,z) \]
Subtract \(R_1\) from \(R_2\):
\[ R_2 \to R_2-R_1 \Rightarrow (-x,y,0) \]
So determinant becomes:
\[ \begin{vmatrix} a & b-y & c-z \\ -x & y & 0 \\ 0 & -y & z \end{vmatrix}=0 \]
Step 3: Expand determinant.
Expanding along first row:
\[ a\begin{vmatrix} y & 0 \\ -y & z \end{vmatrix} -(b-y)\begin{vmatrix} -x & 0 \\ 0 & z \end{vmatrix} +(c-z)\begin{vmatrix} -x & y \\ 0 & -y \end{vmatrix}=0 \]
Step 4: Compute each minor.
\[ \begin{vmatrix} y & 0 \\ -y & z \end{vmatrix}=yz \]
\[ \begin{vmatrix} -x & 0 \\ 0 & z \end{vmatrix}=-xz \]
\[ \begin{vmatrix} -x & y \\ 0 & -y \end{vmatrix}=xy \]
Step 5: Substitute back.
\[ a(yz)-(b-y)(-xz)+(c-z)(xy)=0 \]
\[ ayz + (b-y)xz + (c-z)xy=0 \]
Step 6: Expand.
\[ ayz + bxz - yxz + cxy - zxy = 0 \]
\[ ayz + bxz + cxy - xyz - xyz = 0 \]
\[ ayz + bxz + cxy - 2xyz = 0 \]
Step 7: Divide by \(xyz\).
\[ \frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}-2=0 \]
\[ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2 \]
Final Answer:
\[ \boxed{2} \]