Q80 / 180 · KCET 2008 · Mathematics

If $A=\left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$ and $B=\left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$, then

  • A
    $\mathrm{A}=-\mathrm{B}$
  • Correct answer
    B
    $\mathrm{A}=\mathrm{B}$
  • C
    $\mathrm{B}=0$
  • D
    $B=A^{2}$

Solution

Let $[A]=\left[\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right]$ and $$ \begin{aligned} {[B] } &=\left[\begin{array}{lll} c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right]=\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right] \\ &=[A]^{t} \end{aligned} $$ As we know, $\operatorname{det}()=\operatorname{det}()$ $\therefore \quad \mathrm{A}=\mathrm{B}$

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

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