Let N = left| begin{array}{ccc} 28 & 25 & 38 | vedclass

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(B) First,we evaluate the determinant $N = \left| \begin{array}{ccc} 28 & 25 & 38 \ 42 & 38 & 65 \ 56 & 47 & 83 \end{array} ight|$.Applying row op

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$8$

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(B) First,we evaluate the determinant $N = \left| \begin{array}{ccc} 28 & 25 & 38 \ 42 & 38 & 65 \ 56 & 47 & 83 \end{array} ight|$.Applying row operations: $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$,we get:$N = \left| \begin{array}{ccc} 28 & 25 & 38 \ 14 & 13 & 27 \ 14 & 9 & 18 \end{array} ight|$.Applying $R_2 	o R_2 - R_3$,we get:$N = \left| \begin{array}{ccc} 28 & 25 & 38 \ 0 & 4 & 9 \ 14 & 9 & 18 \end{array} ight|$.Expanding along the first column: $N = 28(4 	imes 18 - 9 	imes 9) - 0 + 14(25 	imes 9 - 38 	imes 4) = 28(72 - 81) + 14(225 - 152) = 28(-9) + 14(73) = -252 + 1022 = 770$.The prime factorization of $N = 770$ is $2 	imes 5 	imes 7 	imes 11$.The number of ways to express $N$ as a product of two relatively prime divisors is given by $2^{n-1}$,where $n$ is the number of distinct prime factors.Here,$n = 4$,so the number of ways is $2^{4-1} = 2^3 = 8$.

Let $N = \left| \begin{array}{ccc} 28 & 25 & 38 \\ 42 & 38 & 65 \\ 56 & 47 & 83 \end{array} \right|$. Then,the number of ways in which $N$ can be expressed as a product of two relatively prime divisors is:

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