If A_1B_1C_1, A_2B_2C_2, A_3B_3C_3 are three- | vedclass

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(A) Let the three-digit numbers be represented as $N_1 = 100A_1 + 10B_1 + C_1 = pk$,$N_2 = 100A_2 + 10B_2 + C_2 = qk$,and $N_3 = 100A_3 + 10B_3 + C_3

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$k$

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(A) Let the three-digit numbers be represented as $N_1 = 100A_1 + 10B_1 + C_1 = pk$,$N_2 = 100A_2 + 10B_2 + C_2 = qk$,and $N_3 = 100A_3 + 10B_3 + C_3 = rk$,where $p, q, r$ are integers.We apply the column operation $C_3 ightarrow C_3 + 10B_2 + 100A_1$ is not directly applicable,but we can use $C_3 ightarrow C_3 + 10C_2 + 100C_1$ is incorrect. Instead,consider the operation $C_3 ightarrow 100C_1 + 10C_2 + C_3$ is not possible,but we can express the determinant by performing $C_3 ightarrow 100C_1 + 10C_2 + C_3$ is not valid. Actually,by properties of determinants,if we perform $C_3 ightarrow 100C_1 + 10C_2 + C_3$,the new third column becomes the values $N_1, N_2, N_3$ which are $pk, qk, rk$.$\Delta = \begin{vmatrix} A_1 & B_1 & 100A_1 + 10B_1 + C_1 \ A_2 & B_2 & 100A_2 + 10B_2 + C_2 \ A_3 & B_3 & 100A_3 + 10B_3 + C_3 \end{vmatrix} = \begin{vmatrix} A_1 & B_1 & pk \ A_2 & B_2 & qk \ A_3 & B_3 & rk \end{vmatrix}$.Taking $k$ common from the third column,we get $\Delta = k \begin{vmatrix} A_1 & B_1 & p \ A_2 & B_2 & q \ A_3 & B_3 & r \end{vmatrix}$.Since the determinant is $k$ times an integer,$\Delta$ is divisible by $k$.

If $A_1B_1C_1, A_2B_2C_2, A_3B_3C_3$ are three-digit numbers,each of which is divisible by $k$,and $\Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}$,then $\Delta$ is divisible by:

Similar Questions

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