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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Determinants

NDA 2025 · Mathematics · Determinants

The value of the determinant vmatrix a and b and c l and m and n p...

The value of the determinant \begin{vmatrix} a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \\ p \text{ and } q \text{ and } r \end{vmatrix} is equal to

Options

  1. A∣a and b and cp and q and rl and m and n∣\begin{vmatrix} a \text{ and } b \text{ and } c \\ p \text{ and } q \text{ and } r \\ l \text{ and } m \text{ and } n \end{vmatrix}
  2. B∣l and m and na and b and cp and q and r∣\begin{vmatrix} l \text{ and } m \text{ and } n \\ a \text{ and } b \text{ and } c \\ p \text{ and } q \text{ and } r \end{vmatrix}
  3. C∣p and q and ra and b and cl and m and n∣\begin{vmatrix} p \text{ and } q \text{ and } r \\ a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \end{vmatrix}
  4. D∣a and p and lb and q and mc and r and n∣\begin{vmatrix} a \text{ and } p \text{ and } l \\ b \text{ and } q \text{ and } m \\ c \text{ and } r \text{ and } n \end{vmatrix}

Correct answer

C. ∣p and q and ra and b and cl and m and n∣\begin{vmatrix} p \text{ and } q \text{ and } r \\ a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \end{vmatrix}

Step-by-step solution

Let Δ=∣a and b and cl and m and np and q and r∣\Delta = \begin{vmatrix} a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \\ p \text{ and } q \text{ and } r \end{vmatrix}

Interchanging the first and third rows (R_1 \leftrightarrow R_3),thesignofthedeterminantchanges:<br><br>Δ=−∣p and q and rl and m and na and b and c∣), the sign of the determinant changes:<br><br>\Delta = -\begin{vmatrix} p \text{ and } q \text{ and } r \\ l \text{ and } m \text{ and } n \\ a \text{ and } b \text{ and } c \end{vmatrix}

Now, interchanging the second and third rows (R_2 \leftrightarrow R_3),thesignchangesagain:<br><br>Δ=−(−∣p and q and ra and b and cl and m and n∣)=∣p and q and ra and b and cl and m and n∣), the sign changes again:<br><br>\Delta = -\left( -\begin{vmatrix} p \text{ and } q \text{ and } r \\ a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \end{vmatrix} \right) = \begin{vmatrix} p \text{ and } q \text{ and } r \\ a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \end{vmatrix}

Answer: ∣p and q and ra and b and cl and m and n∣\begin{vmatrix} p \text{ and } q \text{ and } r \\ a \text{ and } b \text{ and } c \\ l \text{ and } m \text{ and } n \end{vmatrix}
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