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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
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2372 Upvotes
Verified Answer
The correct answer is:
$\mathrm{A}=\mathrm{B}$
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, $\operatorname{det}()=\operatorname{det}()$
$\therefore \quad \mathrm{A}=\mathrm{B}$
$[A]=\left[\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right]$
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, $\operatorname{det}()=\operatorname{det}()$
$\therefore \quad \mathrm{A}=\mathrm{B}$
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