If A=left|begin{array}{ccc}a_{1} & b_{1} & c_ | vedclass

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(B) Given the determinant $A = \left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}\end{array}ight|$.We a

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$A=B$

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(B) Given the determinant $A = \left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}\end{array}ight|$.We are given $B = \left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \ a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3}\end{array}ight|$.First,swap the first and second rows of $B$ to get $B' = -\left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \ c_{1} & c_{2} & c_{3} \ b_{1} & b_{2} & b_{3}\end{array}ight|$.Next,swap the second and third rows of $B'$ to get $B'' = -(-1) \left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \ c_{1} & c_{2} & c_{3}\end{array}ight| = \left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \ c_{1} & c_{2} & c_{3}\end{array}ight|$.Since the determinant of a matrix is equal to the determinant of its transpose,$\left|\begin{array}{ccc}a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \ c_{1} & c_{2} & c_{3}\end{array}ight| = \left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}\end{array}ight| = A$.Therefore,$A = B$.

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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