Divisibility of 19: Bizarre Matrix Problem with Surprising Solution

[Surreal Ike]

I found this odd question in the back of the chapter supplement. I don't need to do it for homework and it probably won't be on the test but the fact that I don't know how to do the problem frustrates me.

Here it is:

* * *

Use the fact that 21,375, 38,798, 34,162, 40,223, and 79,154 are all divisible by 19 to show that

| 2 1 3 7 5 |
| 3 8 7 9 8 |
| 3 4 1 6 2 |
| 4 0 2 2 3 |
| 7 9 1 5 4 |

is divisible by 19 without directly evaluating the determinant.

* * *

It wasn't too hard to notice that the digits of the numbers given were also the digits you get from reading off the numbers in the rows... but this seems like a very superficial relationship!

 

[hamster143]

Do you remember the property that the determinant of a matrix remain constant if you add a multiple of any column to any other column?

 

[Surreal Ike]

No, but now I do. What are you getting at?

 

[hamster143]

Try to add some other columns to the last column.

 

[HallsofIvy]

The point being that 2(10000)+ 1(1000)+ 3(100)+ 7(10)+ 5= 21375, 3(10000)+ 8(1000) +7(100)+ 9(10)+ 8= 38798, etc. That particular linear combination of the numbers on anyone row is divisible by the same number.

 

[hamster143]

Well, if you want to put it *THAT* bluntly...

 

[Surreal Ike]

Thank you.