Question

Let the three digit numbers A28,3B9 and 62C where A, B, and C are any integers between 0 and 9, be divisible by a fixed integer k. Show that the determinant
$open vertical bar table row straight A cell space space 3 end cell cell space space space 6 end cell row 8 cell space 9 end cell cell space space straight C end cell row 2 cell space straight B end cell cell space space 2 end cell end table close vertical bar$
is divisible by k.

Solution

The numbers A28, 3B9, 62C are divisible by k.
Let A28 = ka, 3B9 = k b, 62C = kc where a , b , c are integers.

Let space increment space equals space open vertical bar table row straight A cell space space 3 end cell cell space space 6 end cell row 8 cell space space 9 end cell cell space space straight C end cell row 2 cell space space straight B end cell cell space space 2 end cell end table close vertical bar

equals space open vertical bar table row straight A 3 6 row cell 8 plus 100 straight A plus 10.2 end cell cell space space space space space space space 9 plus 100. space 3 plus 10. space straight B end cell cell space space space space space space space space space straight C plus 100.6 plus 10.2 end cell row 2 straight B 2 end table close vertical bar space space

straight R subscript 2 plus 100 space straight R subscript 1 plus 10 space straight R subscript 3

equals space open vertical bar table row straight A cell space space space space 3 end cell cell space space space space 6 end cell row cell straight A 28 end cell cell space space space 3 straight B 9 end cell cell space space space space 62 straight C end cell row 2 straight B cell space space space 2 end cell end table close vertical bar equals space open vertical bar table row straight A cell space space space 3 end cell cell space space space 6 end cell row cell straight k space straight a end cell cell space space space straight k space straight b end cell cell space space space space straight k space straight c end cell row 2 straight B cell space space 2 end cell end table close vertical bar space equals space straight k space open vertical bar table row straight A cell space space space 3 end cell cell space space space 6 end cell row straight a cell space space straight b end cell cell space space space straight c end cell row 2 cell space space straight B end cell cell space space space 2 end cell end table close vertical bar equals space integral space multiple space of space straight k therefore space space space space space increment space is space divisible space by space straight k.

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