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If $A=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$ and $B=\left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$ then
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2174 Upvotes
Verified Answer
The correct answer is:
$A=B$
We have,
$$
\begin{aligned}
A &=\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right| \\
&=\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right| \\
&=(-)\left|\begin{array}{lll}
b_{1} & b_{2} & b_{3} \\
a_{1} & a_{2} & a_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right| \\
& {\left[\because|A|=\left|A^{T}\right|\right] } \\
&=(-)(-)\left|\begin{array}{lll}
c_{1} & c_{2} & c_{3} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|=\left|\begin{array}{lll}
c_{1} & c_{2} & c_{3} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{2}
\end{array}\right|=B\left[R_{1} \leftrightarrow R_{3}\right]
\end{aligned}
$$
$$
\begin{aligned}
A &=\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right| \\
&=\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right| \\
&=(-)\left|\begin{array}{lll}
b_{1} & b_{2} & b_{3} \\
a_{1} & a_{2} & a_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right| \\
& {\left[\because|A|=\left|A^{T}\right|\right] } \\
&=(-)(-)\left|\begin{array}{lll}
c_{1} & c_{2} & c_{3} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|=\left|\begin{array}{lll}
c_{1} & c_{2} & c_{3} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{2}
\end{array}\right|=B\left[R_{1} \leftrightarrow R_{3}\right]
\end{aligned}
$$
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