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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(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:

Similar Questions

If $\left| \begin{array}{ccc} x - 1 & 3 & 0 \\ 2 & x - 3 & 4 \\ 3 & 5 & 6 \end{array} \right| = 0$,then $x =$

$\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|=$ . . . . . .

The value of the determinant $\left| \begin{array}{ccc} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{array} \right|$ is

If $A = \begin{bmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{bmatrix}$ is a singular matrix and the distinct values of $x$ are $x_1$ and $x_2$,then $x_1 + x_2 + x_1 x_2 = $.

Let $\left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module