Determinant

[epkid08]

Let's say I have a matrix A:

A=\begin{bmatrix}
f(x)& z_1(x)& z_2(x)\\
0& a(x)& b(x)\\
0& c(x)& d(x)
\end{bmatrix}

I've noticed that the determinant of A will either be a(x)d(x) - b(x)c(x) or f(x)a(x)d(x)-f(x)b(x)c(x). I've never found an example of it taking another form. My question is, is there a way to determine which one it is? Does it depend soley on f(x), or does it depend on all functions in the matrix?

[Pengwuino]

Why would f(x) not be included in the determinant?

[epkid08]

Why would f(x) not be included in the determinant?

I'm not sure why, but I've found an example, I'll post it in a minute.

[HallsofIvy]

The 2 by 2 determinant
\left|\begin{array}{cc}a & b \\c & d\end{array}\right|
is defined as "ad- bc".

The 3 by 3 determinant
\left|\begin{array}{ccc}a & b & c\\d & e & f\\ g & h & i\end{array}\right|[/itex]
can be "expanded by the first column" to give
[tex]a\left|\begin{array}{cc}e & f \\ h & i\end{array}\right|- d\left|\begin{array}{cc}b & c \\ h & i\end{array}\right|+ g\left|\begin{array}{cc}b & c \\ e & r\end{array}\right|
= a(ei- fh)- d(bi- ch)+ g(bf- ce) .

In the special case that the bottom two entries of the first column are 0, that reduces to what you have:
a\left|\begin{array}{cc}e & f \\ h & i\end{array}\right|
= a(ei- fh).

[epkid08]

Hmm, it's very possible that my book made a typo.

This is the example from my book:

Find the Wronskian of the set \{x,x^2,x^3\}

W(x, x^2, x^3)=\left|\begin{array}{ccc}x & x^2 & x^3 \\1 & 2x & 3x^2\\ 0 & 2 & 6x\end{array}\right|

xR_2-R_1 \rightarrow R_2

W(x, x^2, x^3)=\left|\begin{array}{ccc}x & x^2 & x^3 \\0 & x^2 & 2x^3\\ 0 & 2 & 6x\end{array}\right|

Now according to you, the Wronskian should evaluate to x((x^2)(6x)-(2)(2x^3))=2x^4, but the book has the answer as 2x^3.

[epkid08]

Plugging in x=5 into the function, and using a calculator to evaluate the determinant, it seems like it was a typo all along.

Ehh! :yuck:

[epkid08]

Another question.

So if I had a n by n matrix B:


B_{n,n} =
\begin{bmatrix}
f_1(x) & a_2 & \cdots & a_{n-1}& a_n \\
0 & f_2(x) & \cdots & b_{n-1}&b_n \\
\vdots & \vdots & \ddots & \vdots&\vdots \\
0 & 0 &\cdots &f_{n-1}(x) & z_n\\
0 & 0 & \cdots & 0&f_n(x)
\end{bmatrix}
\]


(Ignoring my misuse of Latex, hopefully you know what I mean)

Does \det(B) always equal \prod^n_{k=1}f_k(x)?

[Pengwuino]

Does \det(B) always equal \prod^n_{k=1}f_k(x)?

Yes, if every entry below the diagonal is 0, then the determinant is simply the product of every entry in the diagonal, irregardless of what is going on above the diagonal.

[Dafe]

It follows from the fact that det I = 1.
By eliminating entries above the pivots in your upper triangular matrix, it can be made into a diagonal matrix.

Since we also know that the determinant is a linear function of each row/column separately, we may factor out:
a_{i,j} for i = j you get:

a_{11}a_{22}...a_{nn}I

As mentioned, det I = 1 and what you get is a product of every entry in the diagonal, just as mr. Pengwuino said.

I am just learning this stuff myself, hope I have not written something crazy :)

[HallsofIvy]

Or given that A is upper triangular, expand det(A) by the first diagonal:
\left|\begin{array}{cccc}a & b & ... & c \\ 0 & d & ... & e \\ ... & ... & ... & ... \\ 0 & 0 & ... & z\end{array}\right|= a\left|\begin{array}{ccc}d & ... & e \\ ... & ... & ... \\ 0 & 0 & z\end{array}\right|
and expand that determinant by the first diagonal, etc.