Q44 / 160 · AP EAMCET 2015 · Mathematics

The value of the determinant $\left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\ a b-a^2 & a-b & b^2-a b \\ b c-a c & c-a & a b-a^2\end{array}\right|$ is

  • A
    $abc$
  • B
    $a+b+c$
  • Correct answer
    C
    $0$
  • D
    $ab + bc + ca$

Solution

$\begin{aligned} & \text { Let } \mathrm{D}=\left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\ a b-a^2 & a-b & b^2-a b \\ b c-a c & c-a & a b-a^2\end{array}\right| \\ & =\left|\begin{array}{lll}b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a)\end{array}\right| \\ & =(b-a)(b-a)\left|\begin{array}{lll}b & b-c & c \\ a & a-b & b \\ c & c-a & a\end{array}\right|\end{aligned}$ $=(b-a)^2\left|\begin{array}{ccc}b & 0 & c \\ a & 0 & b \\ c & 0 & a\end{array}\right| \quad \quad\left[\operatorname{using} C_2 \rightarrow C_2-\left(C_1-C_3\right)\right]$ $=0$

Hence, the correct answer is option: 'C'.

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