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NDA 2 2025 Maths, question 19

Topic: Algebra

Q19

If a,b,ca,b,c are the sides of a triangle ABCABC and pp is the perimeter of the triangle, then what is

∣p+cabcp+abcap+b∣\left| \begin{matrix} p+c & a & b \\ c & p+a & b \\ c & a & p+b \end{matrix} \right|

equal to?

  1. (a) p3p^{3}
  2. (b) 2p32p^{3}Correct answer
  3. (c) 3p33p^{3}
  4. (d) 4p34p^{3}

Solution.

Answer: (b) 2p32p^{3}

  1. Subtract row 3 from rows 1 and 2 to obtain (p,0,−p) and (0,p,−p).
  2. Factor p from each of these rows, then add columns 1 and 2 to column 3.
  3. Its only nonzero entry becomes c+a+p+b=2p.
  4. Hence the determinant is p²×2p=2p³.