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.
