For any \(a, b, c \in \mathbf{R}\), the determinant \(\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1…

For any \(a, b, c \in \mathbf{R}\), the determinant \(\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|\) is equal to
  1. \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
  2. \(a(b-c)+b(c-a)+c(a-b)\)
  3. \((a-b)(b-c)(c-a)\)
  4. \(a b c\)

Solution

For any \(a, b, c \in \mathbf{R}\), the given determinant \(\Delta=\left|\begin{array}{lll} b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1 \end{array}\right|\) On applying \(R_2 \rightarrow R_2-R_1\) and \(R_3 \rightarrow R_3-R_1\), we have \(\begin{gathered} \Delta=\left|\begin{array}{ccc} b c & b+c & 1 \\ c(a-b) & a-b & 0 \\ b(a-c) & a-c & 0 \end{array}\right| \\ =(a-b)(a-c)\left|\begin{array}{ccc} b c & b+c & 1 \\ c & 1 & 0 \\ b & 1 & 0 \end{array}\right| \end{gathered}\) On applying \(R_3 \rightarrow R_3-R_2\), we have \(\begin{aligned} \Delta & =(a-b)(a-c)\left|\begin{array}{ccc} b c & b+c & 1 \\ c & 1 & 0 \\ b-c & 0 & 0 \end{array}\right| \\ & =(a-b)(a-c)[0-(b-c)]=(a-b)(b-c)(c-a) \end{aligned}\) Hence option (c) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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