Determinant - 3 X 3

[cepheid]

What's the best way to take the determinant of a 3 X 3 matrix. It's actually a matrix in the form:

\mathbf{A} - \lambda\mathbf{I}

So I figured Gaussian elimination would be ugly, because of all the lambdas floating around. I tried the method of expansion by cofactors...and ended up with the cubic characteristic polynomial, only to find I had no idea how to solve it. I took that to mean that I had done something wrong, so rather than posting that result, I thought I'd get some advice on how to start from scratch. The matrix is symmetric, btw, if that helps the situation.

[Tide]

You won't get around the fact that the determinant is cubic in \lambda. I presume you want eventually to set the determinant to zero and solve for \lambda. You may or may not be able to factor the cubic equation but you have graphing and other approximation methods at your disposal.

[shmoe]

Usually a combination of row (or column) operations and cofactor expansion is the way to go. Do some convenient row operations to make a row or column of mostly zeros and expand. It's tough to say what is the least error prone method, it depends on where you want to do your work and what your matrix actually looks like.

There are formulas for solving cubics by the way, though if this came up in a linear algebra course, usually they're cooked to have at least one easy to find root.

[dextercioby]

You won't get around the fact that the determinant is cubic in \lambda. I presume you want eventually to set the determinant to zero and solve for \lambda. You may or may not be able to factor the cubic equation but you have graphing and other approximation methods at your disposal.

Try to use Girolamo Cardano's formulas.They are 500 years old and apparently the ignorant people forgot about them.Nobody's asking u to memorize them,just to look'em up in an (very) old fashioned algebra book. :wink:

Daniel.

[Tide]

Dexter,

Of course, but if you've actually ever tried to USE Cardano's formula you'll appreciate numerical approximation! :-)

On the other hand, Derive (e.g.) does a pretty good job of handling cubic equations so you can avoid the tedium.

[dextercioby]

Dexter,

Of course, but if you've actually ever tried to USE Cardano's formula you'll appreciate numerical approximation! :-)

On the other hand, Derive (e.g.) does a pretty good job of handling cubic equations so you can avoid the tedium.

Well,i'm not to keen at handling computer software (but i know how to use a pocket calculator :wink: ),so any numerical solutions with help of computer techniques,for me,is refuted from the start.

Yes,i'd take that old dusty book from the bookshelf and search for Cardano's formulas,than wonder how to get an approximate solution,when an exact one is available.

As for numerical approximation using the CPU on the shoulders,sometimes mine just fails when encounteriing something like that
12\sqrt{7} x^{3}+4\sqrt [3] {5} x^{2}-5x+8=0 :tongue2:

Daniel.

[Tide]

Daniel,

You misunderstood. A package like Derive isn't intended to give you an approximation - it's a symbolic algebra package which has the advantage of giving you the result of using Cardano's formula with all the coefficients substituted in appropriately. If you have used a pocket calculator then you know it gets a little hairy.

BTW - this is what Derive gives for your problem

x = 0.09800530325 - 0.3839643014· ∨ x = 0.09800530325 + 0.3839643014· ∨ x = -1.604601031

which came from the EXACT solutions:

x = (- √(- 125·√7/146313216 + 625·√5/3429216 + 25·√35/1016064 + 105739/438939648) + 625·√35/285768 + 25·√5/12096 + √7/168)^(1/3) + (√(- 125·√7/146313216 + 625·√5/3429216 + 25·√35/1016064 + 105739/438939648) + 625·√35/285768 + 25·√5/12096 + √7/168)^(1/3) - 5·√35/63 + ·(√3·(√(- 125·√7/2286144 + 1250·√5/107163 + 25·√35/15876 + 105739/6858432) + 625·√35/35721 + 25·√5/1512 + √7/21)^(1/3)/2 - √3·(- √(- 125·√7/2286144 + 1250·√5/107163 + 25·√35/15876 + 105739/6858432) + 625·√35/35721 + 25·√5/1512 + √7/21)^(1/3)/2) ∨ x = (- √(- 125·√7/146313216 + 625·√5/3429216 + 25·√35/1016064 + 105739/438939648) + 625·√35/285768 + 25·√5/12096 + √7/168)^(1/3) + (√(- 125·√7/146313216 + 625·√5/3429216 + 25·√35/1016064 + 105739/438939648) + 625·√35/285768 + 25·√5/12096 + √7/168)^(1/3) - 5·√35/63 + ·(√3·(- √(- 125·√7/2286144 + 1250·√5/107163 + 25·√35/15876 + 105739/6858432) + 625·√35/35721 + 25·√5/1512 + √7/21)^(1/3)/2 - √3·(√(- 125·√7/2286144 + 1250·√5/107163 + 25·√35/15876 + 105739/6858432) + 625·√35/35721 + 25·√5/1512 + √7/21)^(1/3)/2) ∨ x = - (- √(- 125·√7/2286144 + 1250·√5/107163 + 25·√35/15876 + 105739/6858432) + 625·√35/35721 + 25·√5/1512 + √7/21)^(1/3) - (√(- 125·√7/2286144 + 1250·√5/107163 + 25·√35/15876 + 105739/6858432) + 625·√35/35721 + 25·√5/1512 + √7/21)^(1/3) - 5·√35/63

which I won't bother TeXifiying! (It shows up perfectly rendered on the Derive screen. :-)

[cepheid]

Thanks for the help guys, it's much appreciated. I'm getting some more practice with 3 x 3 systems. I knew that the characteristic polynomial would have to be cubic, but just as somebody mentioned, I was expecting it to be easy to factor, and it wasn't. I suppose they could give us a hard one, but given the number of problems in the homework, I don't think they were expecting us to go out and get "Derive" or spend half a day looking up formulas in some musty tome ;)

IIRC, such formulas do not exist and cannot for polynomials of degree 4 and higher. What then...?

[theFuture]

cepheid: Have you tried synthetic division and the rational root theorem? That's usually where I go first when I have trouble factoring by inspection.

[dextercioby]

IIRC, such formulas do not exist and cannot for polynomials of degree 4 and higher. What then...?

It's a certain fact known in mathematics that equations for polynomials with real/complex coeffcients can be solved only for degrees smaller than 5.For the degrees 5,6,7,...,there is no general method,and in most cases,no factorization can occur.

I believe that in the "musty tome",the next page after dealing with Girolamo Cardano's formulas for the third degree algebraic equation,there is the general method (involving simple substitutions) in which ANY EQUATION OF THE 4-th order can be transformed into one of the 3-rd order and solved using Girolamo Cardano's formulas.

Daniel.

PS.The history involving three italian dudes:Scipione del Ferro,Niccolo Tartaglia and Girolamo Cardano and the famous "cubic formulas" is among the most famous in the history of mathematics.As famous as the "Delic problems". :wink: