# Answer to Question #12455 in Linear Algebra for john.george.milnor

Question #12455

Show that determinant

2 0 6 0 4

5 3 2 2 7

2 5 7 5 5

2 0 9 2 7

0 0 2 8 9

is divisible by 17

2 0 6 0 4

5 3 2 2 7

2 5 7 5 5

2 0 9 2 7

0 0 2 8 9

is divisible by 17

Expert's answer

The determinant will have no changes after adding one column to another,

multiplied by some number.

So we can

1)multiply first column by 10000 and

add to the last one

2)multiply second column by 1000 and add to the last

one

3)multiply third column by 100 and add to the last one

4)multiply

forth column by 10 and add to the last one

After this elementary column

operations last column will be

20604

53227

25755

20927

289

All these numbers are divisible

by 17, so as one column is divisible by 17,

then the whole determinant is

divisible by 17 too.

multiplied by some number.

So we can

1)multiply first column by 10000 and

add to the last one

2)multiply second column by 1000 and add to the last

one

3)multiply third column by 100 and add to the last one

4)multiply

forth column by 10 and add to the last one

After this elementary column

operations last column will be

20604

53227

25755

20927

289

All these numbers are divisible

by 17, so as one column is divisible by 17,

then the whole determinant is

divisible by 17 too.

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