number grid

Einsteins,

Can you please help me with number grid (GCSE) maths coursework?

Regards
Khushi

Yes. The answer is 3.

I should add: you need to give more information than that to get help as many people don't know the ins and outs of that particular coursework.

Reply 2

vector

Do you have a brief or anything?

Reply 3

sheoran

thanks for your response. The grid is of 10X10 i.e. 12345678910 11121314151617181920...

- a box is drawn round four numbers
- find the product of the top left number and the bottom right number in this box
- do the same with top right and bottom left numbers
- calculate the difference between these products

investigate further

Reply 4

zrancis

Okay, so have you noticed a relationship between the number you get and where the box is or any other feature?

Post where you have gotten upto, we like to help but you won't be getting a 25 page perfect solution :biggrin:

Reply 5

vector

Try extending the box to 3x3 perhaps. It's a bit beyond GCSE, but by doing your 2x2 thing you are actually finding the determinant of the 2x2 matrix (that is the group of four numbers in your box)

Reply 6

sheoran

i got one relationship for 2x2 i.e.

(product of Top Left and Bottom right) - (product of top right and bottom left) = 10

can u guys pl help for other such relationships or formulas?

Reply 7

vector

Take your box to be 3x3

Find the determinant of this by:

Choosing the top left number. Cross out the row and column it's in. Take the 2x2 matrix left and find it's determinant (as you did before). Multiply this by the top left number you crossed out
Choose the middle top number. Cross out its row and column. Find the determinant of the 2x2 matrix left. Multiply it by the top middle number you crossed out. Do the same for the top right number.

I have done this and the result is of some significance.

Good luck!

Reply 8

sheoran

guys,

at i got one formula for 3x3 :-

sum of cells is y^2*[centre cell]. where y is matrix size i.e 3 in this case.

any idea how is derived?

Reply 9

zrancis

hmm,

Let the matrix start in the top left with number n i.e.

---n-----n+1----n+2
10n+1--10n+2--10n+3
20n+1--20n+2--20n+3

so, the sum of the cell:

n + (n+1) + (n+2) + ... = 93n + 15

But your formula says the sum is:

3^2*(10n+2) = 90n + 18

Writing the matrix out like this can be useful, try letting different starting numbers = n and look for relationships.

Like the determinant of a 2*2:

---n-----n+1
10n+1--10n+2

Det= n(10n+2) - (n+1)(10n+1)
=10n^2 + 2n - 10n^2 -11n -1
=-9n - 1

So to find the determinant of a 2*2 matrix, multiply the first term by -9 and take 1

Reply 10

vector

sheoran

guys,

at i got one formula for 3x3 :-

sum of cells is y^2*[centre cell]. where y is matrix size i.e 3 in this case.

any idea how is derived?

I'll have a look now.

Note also that the determinant of the 3x3 matrix is always 0.

Reply 11

sheoran

vector,
how did u derive that determinant of 3x3 matrix is zero?

Reply 12

vector

Take your box to be 3x3

Find the determinant of this by:

Choosing the top left number. Cross out the row and column it's in. Take the 2x2 matrix left and find it's determinant (as you did before). Multiply this by the top left number you crossed out
Choose the middle top number. Cross out its row and column. Find the determinant of the 2x2 matrix left. Multiply it by the top middle number you crossed out. Do the same for the top right number.

The proof of this is beyond even my abilities, so will not be required for your GCSE coursework :wink: