Problem Analysis: The question asks for the value of $k$ given an equation involving two determinants. The determinant on the left involves differences between elements, while the one on the right is a standard determinant.

Determinant Value k Calculation

Let the given equation be:

$ D_1 = \begin{vmatrix} a-b & p-q & x-y \\ b-c & q-r & y-z \\ c-a & r-p & z-x \end{vmatrix} = k \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} = k \cdot D_2 $

Simplify Left-Side Determinant (D₁)

We can use row operations to simplify $D_1$. Apply the operation $R₁ → R₁ + R₂ + R₃$.

• The first element of the new row 1 becomes: $(a-b) + (b-c) + (c-a) = 0$
• The second element becomes: $(p-q) + (q-r) + (r-p) = 0$
• The third element becomes: $(x-y) + (y-z) + (z-x) = 0$

After this operation, the determinant $D_1$ becomes:

$ D_1 = \begin{vmatrix} 0 & 0 & 0 \\ b-c & q-r & y-z \\ c-a & r-p & z-x \end{vmatrix} $

Evaluate Determinant D₁

A fundamental property of determinants is that if a matrix has a row (or column) consisting entirely of zeros, its determinant is zero.

Therefore, $D_1 = 0$.

Solve for k

Substitute the value of $D_1$ back into the original equation:

$0 = k \cdot D_2$

Where $D_2 = \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$.

For this equation to hold true generally (assuming $D_2$ is not necessarily zero), the value of $k$ must be 0.

Conclusion

The value of $k$ is 0.