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

• epkid08

#### 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?

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

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.

The 2 by 2 determinant
$$\left|\begin{array}{cc}a & b \\c & d\end{array}\right|$$

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).

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 $$2x^3$$.

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:

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 $$\prod^n_{k=1}f_k(x)$$?

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.

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 :)

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.