Evaluating 4 X 4 determinants?

Question

I remember from class that we learned how to do these but there were cases where you had to use cramer's rule. anyways...I'm stuck on this one because I expanded it and then I didn't know what to do. I can't find any examples like that in my notes. I need to know what to do next or if I should have used the dx, dy, dz thing. the problem is

a 0 2 1
b 1 3 0
c 0 1 2
d 1 0 1

thanx!!!!!!!!!!!

Answer 1

Are a, b, c, and d variables? If so,

Determinant = a ((1)(1-0) - (0) + (1)(6 - 0)) - b ( 0 - 0 + (1)(1 -4)) + c (0 - (1)(2 - 0) + (1)(0 - 3)) - d (0 - (1)(4 - 1) + 0)

= a (7) - b(-3) + c(-5) - d(-3)
= 7a + 3b -5c + 3d <<< answer

And no, you don't need to do any calculus in solving for the determinant. You only need Cramer's rule if you are evaluating a system of linear equations.

Answer 2

Expand by minors. Break it into four 3 by 3 determinants. Use the second column because it has two zeroes in it. The lead coefficient above each of the 3 by 3 determinants is the scalar by which the determinant should be multiplied. In the case of a lead coefficient of zero, I didn't bother showing the determinant, since you just get zero when you multiply it by zero.

4 by 4 Det =

-0

+1
a 2 1
c 1 2
d 0 1

-0

+1
a 2 1
b 3 0
c 1 2

You'll note that we have broken it down to the sum of two 3 by 3 determinants. If you don't know how to solve a 3 by 3 determinant directly, repeat the process of expanding by minors.

Answer 3

...| 1 3 0 |.....| b 2 1 |.....| b 1 0 |.....| b 1 3 |
a.| 0 1 2 | -0.| c 3 0 | +2.| c 0 2 | -1.| c 0 1 |
...| 1 0 1 |.....| d 0 1 |.....| d 1 1 |.....| d 1 0 |

In other words, to go from NxN to (N-1)x(N-1), take each term from the top row multiplied by +1 for odd and -1 for even positions, and multiply that value by the rows below it and the columns that don't contain it.

Answer 4

cofactor expansion
determinant =
a*(-1)^(1+1) * det(1 3 0) - 0 + 2*(-1)^(1+3)*det(b 1 0)
(0 1 2) (c 0 2)
(1 0 1) (d 1 1)
+ 1*(-1)^(1+4) * det(b 1 3)
(c 0 1)
(d 1 0)
= a[(1)(1)(1) + (3)(2)(1) + 0 - 0 - 0 - 0] - 2[0 + (1)(2)(d) + 0 - (1)(2)(b) - (1)(c)(1)] - 1[0 + (1)(1)(d) + (3)(c)(1) - 0 - (1)(1)(b) - 0]
= 7a - 4d + 4b +2c - d - 3c +b
= 7a + 5b - c - 5d

Answer 5

a11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 a41 a42 a43 a44 a11*a22*a33*a44 + a12*a23*a34*a41 + a13*a24*a42*a31 + a14*a43*a32*a21 - a41*a32*a23*a14 - a31*a22*a13*a44 - a21*a12*a34*a43 - a11*a24*a33*a42 = D

Answer 6

i'm learning the same thing but haven't got that in depth yet