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If \(\begin{vmatrix} a-b & p-q & x-y \\ b-c & q-r & y-z \\ c-a & r-p & z-x \end{vmatrix} = k \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}\) then what is the value of k ?

This question was previously asked in
NDA 2 2026 GAT Question Paper (13-Sep-2026)
The correct answer is
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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.

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