matrices: calculate the determinant?

Question

im having a problem with matrices. This is one question:

| 2 0 3 |
| -1 1 1 |
| 0 4 1 |

it says, "calculate the determinants first by an expansion on the 1st row and then by an expansion on the 3rd row. in each case verify that the two expansions give you the same answer"

i have the solution infront of me, but it is not clear how they came about with the answer.

can someone kindly post a step by step explanation please? your help is greatly appreciated! =)

Answer 1

1st row ...2 det of 1,1 over 4,1 + 3 det of -,1 over 0,4 = 2 [1-4] + 3 [ -4 -0]



3rd row -4 det of 2,3 over -1,1 + 1 det of 2,0 over -1,1 = -4 [2 -(-3)] + 1 [2 - 0]

Answer 2

http://www.youtube.com/watch?v=CLuxS_nb5Ik

the above video helps a lot

you can find value of determinant by expamding at any row or column

important part is the i+ 1 bvalue which idicated the +1 or - 1 multiplier

Now row1 col 1 (r1c1) means i = 1and j = 1 or i + j = 1+1 =2 here (-1) ^ i+ j = (-1) 2 = +1 is muliplier

or row1 col 2 (r1,c2)means i = 1and j = 2 or i + j = 1+2 =3 here (-1) ^ i+ j = (-1) 3 = -1 is muliplier

Now it is easy suppose you want to go by first row

in your determinant



first we have 2 and with +1 as multiplier as shown above for row1 col 1

now hide the row and column of r1 c1

and calculate value of detrminant or (1*1 - 1*4) in our case

so we have (+1) 2(1*1 - 1*4)

now we move to r1c2 which has nagative -1 as multipier

our number is 0 and hiding row1col 2 we have (-1*1- 1*0) so we get (-1)* 0 * (-1*1- 1*0)

we do this for r1 c3 where multipier is + 1 and we have (+1)(3) (-1*4 -0*1)

determinant value is (+1) *2*(1*1 - 1*4)+ (-1)* 0 * (-1*1- 1*0) + (+1)(3) (-1*4 -0*1)

the above is value of determinant



if we expand by row 3

we go thru same steps

take r3c1 with multiplier as (-1) ^ (3+1) or +1 we have (+1)(0) (0*1-1*3) obtained by hiding r3col1

then take r3c2 and get (-1) (4) [(2)*(1) - (-1)*3]

and take r3c2 which gives (+1)(1) ( -1*4-0*1)

(+1)(0) (0*1-1*3) + (-1) (4) [(2)*(1) - (-1)*3] + (+1)(1) ( -1*4-0*1)

adding all three we get the naswer

Answer 3

Wow a typo??????? No wonder it was so hard.. A Moore matrix is a matrix over a finite field. So that does not solve your original question (with the typo). Now with the typo fixed... your matrix is just a Vandermonde matrix, since a_(i,j) = j^(2*(i-1)) = (j^2)^(i-1).