Q169 / 270 · NDA 2025 (Phase 2) · Mathematics
If $a$, $b$, $c$ are the sides of a triangle $ABC$ and $p$ is the perimeter of the triangle, then what is $\begin{vmatrix} p+c & a & b \\ c & p+a & b \\ c & a & p+b \end{vmatrix}$ equal to?
- A$p^3$
- Correct answerB$2p^3$
- C$3p^3$
- D$4p^3$
Solution
Let $\Delta = \begin{vmatrix} p+c & a & b \\ c & p+a & b \\ c & a & p+b \end{vmatrix}$
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
$\Delta = \begin{vmatrix} p+a+b+c & a & b \\ p+a+b+c & p+a & b \\ p+a+b+c & a & p+b \end{vmatrix}$
Since $p = a+b+c$ is the perimeter of the triangle, we have $p+a+b+c = 2p$. Substituting this into the determinant:
$\Delta = \begin{vmatrix} 2p & a & b \\ 2p & p+a & b \\ 2p & a & p+b \end{vmatrix}$
Taking $2p$ common from the first column $C_1$:
$\Delta = 2p \begin{vmatrix} 1 & a & b \\ 1 & p+a & b \\ 1 & a & p+b \end{vmatrix}$
Applying the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = 2p \begin{vmatrix} 1 & a & b \\ 0 & p & 0 \\ 0 & 0 & p \end{vmatrix}$
Expanding the determinant along the first column $C_1$:
$\Delta = 2p [1(p \cdot p - 0)] = 2p(p^2) = 2p^3$
Answer: $2p^3$
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
$\Delta = \begin{vmatrix} p+a+b+c & a & b \\ p+a+b+c & p+a & b \\ p+a+b+c & a & p+b \end{vmatrix}$
Since $p = a+b+c$ is the perimeter of the triangle, we have $p+a+b+c = 2p$. Substituting this into the determinant:
$\Delta = \begin{vmatrix} 2p & a & b \\ 2p & p+a & b \\ 2p & a & p+b \end{vmatrix}$
Taking $2p$ common from the first column $C_1$:
$\Delta = 2p \begin{vmatrix} 1 & a & b \\ 1 & p+a & b \\ 1 & a & p+b \end{vmatrix}$
Applying the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = 2p \begin{vmatrix} 1 & a & b \\ 0 & p & 0 \\ 0 & 0 & p \end{vmatrix}$
Expanding the determinant along the first column $C_1$:
$\Delta = 2p [1(p \cdot p - 0)] = 2p(p^2) = 2p^3$
Answer: $2p^3$
Hence, the correct answer is option: 'B'.