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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Determinants

NDA 2026 · Mathematics · Determinants

If vmatrix a-b and p-q and x-y b-c and q-r and y-z c-a and r-p and...

If ∣a−b and p−q and x−yb−c and q−r and y−zc−a and r−p and z−x∣=k∣a and b and cp and q and rx and y and z∣\begin{vmatrix} a-b \text{ and } p-q \text{ and } x-y \\ b-c \text{ and } q-r \text{ and } y-z \\ c-a \text{ and } r-p \text{ and } z-x \end{vmatrix}=k\begin{vmatrix} a \text{ and } b \text{ and } c \\ p \text{ and } q \text{ and } r \\ x \text{ and } y \text{ and } z \end{vmatrix}, then what is the value of kk ?

Options

  1. A−1-1
  2. B00
  3. C12\dfrac{1}{2}
  4. D11

Correct answer

B. 00

Step-by-step solution

Let Δ=∣a−b and p−q and x−yb−c and q−r and y−zc−a and r−p and z−x∣\Delta = \begin{vmatrix} a-b \text{ and } p-q \text{ and } x-y \\ b-c \text{ and } q-r \text{ and } y-z \\ c-a \text{ and } r-p \text{ and } z-x \end{vmatrix}

Applying the row operation R1→R1+R2+R3R_1 \rightarrow R_1 + R_2 + R_3, we get:

Δ=∣(a−b)+(b−c)+(c−a) and (p−q)+(q−r)+(r−p) and (x−y)+(y−z)+(z−x)b−c and q−r and y−zc−a and r−p and z−x∣\Delta = \begin{vmatrix} (a-b)+(b-c)+(c-a) \text{ and } (p-q)+(q-r)+(r-p) \text{ and } (x-y)+(y-z)+(z-x) \\ b-c \text{ and } q-r \text{ and } y-z \\ c-a \text{ and } r-p \text{ and } z-x \end{vmatrix}

Δ=∣0 and 0 and 0b−c and q−r and y−zc−a and r−p and z−x∣\Delta = \begin{vmatrix} 0 \text{ and } 0 \text{ and } 0 \\ b-c \text{ and } q-r \text{ and } y-z \\ c-a \text{ and } r-p \text{ and } z-x \end{vmatrix}

Since all elements of the first row are zero, the value of the determinant is 00.

Comparing this with k∣a and b and cp and q and rx and y and z∣k\begin{vmatrix} a \text{ and } b \text{ and } c \\ p \text{ and } q \text{ and } r \\ x \text{ and } y \text{ and } z \end{vmatrix}, we get k=0k = 0.
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