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

[epkid08]

Let's say I have a matrix A:

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

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
[tex]\left|\begin{array}{ccc}a & b & c\\d & e & f\\ g & h & i\end{array}\right|[/itex]<br /> can be "expanded by the first column" to give<br /> $$a\left|\begin{array}{cc}e & f \\ h & i\end{array}\right|- d\left|\begin{array}{cc}b & c \\ h & i\end{array}\right|+ g\left|\begin{array}{cc}b & c \\ e & r\end{array}\right|$$<br /> = a(ei- fh)- d(bi- ch)+ g(bf- ce) .<br /> <br /> In the special case that the bottom two entries of the first column are 0, that reduces to what you have: <br /> $$a\left|\begin{array}{cc}e & f \\ h & i\end{array}\right|$$<br /> = a(ei- fh).[/tex]

 

[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!

 

[epkid08]

Another question.

So if I had a n by n matrix B:

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

(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, regardless 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.