Math: Matrices/Determinants.......and all that stuff!!?

Question

I have no idea how to work this out!???! I looked through the book, and tried to figure out how to work it, but I can't.Sooo, please help.... Here it is:

Transform the determinant into one in which column 1 contains all zero elements but one, and expand the transformed determinant by this column. Please show your work, so I can learn something..=].

Matrix:
1.....6.....1
-6....-47....-2
-6....-41....2



Thank You!

Answer 1

(1) add row#1*6 from row#2, then from row#3
that is [–6+1*6, –47+6*6, -2+1*6] for row#2
that is [-6+1*6, -41+6*6, 2+1*6] for row#3
thus:
1……6……1
0…..-11…..4
0……-5…..8
(2) add row#2*(-5/11) from row#3
that is [0, -5+(-5)*(-5/11), 8+4*(-5/11)] for row#3
thus:
1……6……1
0…...-11…..4
0……0….68/11
to check the result multiply main diagonal: 1*(-11)*(68/11)=-68

Answer 2

Use row operations to zero out a21 and a31 (by mult row 1 by 6 and adding to each of rows 2 and 3)

Now you have a matrix with a11 = 1 and a21=a31=0

Then you expand the determinant along that column and since only a11 is non-zero, the determinant is

det of the reduced matrix of
a22 a23
a32 a33

Note that these have been changed from the originals by the row operations.

You really should be able to find this in your text.

Answer 3

Here is the value of the second-order determinant:
|a b|
|c d|
= ad - bc

Example:
|1 2|
|3 4|
= (1)(4) - (2)(3)
= 4 - 6
= -2

____________________
Well, you need to know this if you want to know the value of the third-order determinant:
|a b c|
|d e f|
|g h i|
= a·|e f| - b·|d f| + c·|d e|
. . . |h i| . . . |g i| . . . |g h|

Example:
|1 2 3|
|4 5 6|
|7 8 9|

= 1|5 6| - 2|4 6| + 3|4 5|
. .. |8 9| . .. |7 9| . . .|7 8|

= 1[(5)(9) - (6)(8)] - 2[(4)(9) - (6)(7)] + 3[(4)(8) - (5)(7)]
= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)
= 1(-3) - 2(-6) + 2(-3)
= -3 + 12 - 6
= 3
______________
To solve your problem, you may want to find the value of the determinant first, then try to think of values for
|a b c|
|0 d e|
|0 f g|

^_^

Answer 4

They wrote cos^2 (3x) as a function of cos(6x) cos^2 (3x) = a million/2 *[cos(6x) + a million] it extremely is how they have been given the above answer. 2) 10x^4 dx / (x^5 + a million) it extremely is taken into consideration needed to jot down the dx once you're doing those integrals. It skill something and it enables plenty. Use the substitution u = x^5 + a million du = 5x^4 dx So your venture will become 2*du / u integrated you get 2ln(u) = 2*ln(x^5 + a million) + consistent you may additionally write it extra elegantly, ln [C*(x^5 + a million)^2]

Answer 5

my answer is in this link I hope this could help
http://img246.imageshack.us/img246/5449/3154yi6.png