# Basic Multivariable Proof Equating Determinant to dot/cross product

Hello folks! Just a concise introduction of myself before I get to the task at hand: I'm new to these forums although I have been surfing them frequently for the past 5 years! I am not a math major and quite frankly, my skills in the subject are limited. Be that as it may, my fascination for math has inspired me to pursue a career in either pure math or computational science. My name is Mark and I'm an undergrad Uni student.

K, now on to the task at hand: I am not familiar with symbolic/notation input on these forums so I apologize in advance for my crude way of displaying the question in a pictorial format. Simply put, my goal in the picture is to equate the determinant of vectors A, B and C to the dot product of vector A and (vector B cross-product vector C).

Wow that sounded stupid! Lol, please excuse my naïveté. I have a STRONG feeling I am doing something incorrectly. Can anyone please steer me in the right direction?:

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lurflurf
Homework Helper
$$\mathop{det} ( \vec{A} , \vec{B} , \vec{C} )=\vec{A} \cdot ( \vec{B} \times \vec{C})$$

as there is only one function that has the properties each side has, they must be equal.
This could be shown different ways, but I like.
$$\mathop{det} ( \tau \vec{A} , \tau \vec{B} , \tau \vec{C} )=\mathop{det} ( \vec{A} , \vec{B} , \vec{C} )$$
and
$$\mathop{det} ( \vec{i} , \vec{j} , \vec{k} )=1$$
where $$\tau$$ is any rotation
These properties determine det uniquely
since $$\vec{A} \cdot ( \vec{B} \times \vec{C})$$ has these properties as well the two are equal.

lurflurf
Homework Helper
It looks like you are trying to prove it using the coordinate formula, that (while crude) will work also.

Well, blimey, that was simple enough! Looks like what I did was overkill

Unfortunately, I did not know that $\tau$ stood for rotation.

Fredrik
Staff Emeritus
Gold Member
Welcome to PF, Mark. Five years is a long to time to lurk without posting.

##\tau## is not a standard notation for rotation. I think R is the most common notation, but I'm not sure if it can be called "standard". Here I guess ##\tau## looks prettier, since you're using uppercase Latin letters for vectors.

Your own approach works, but the calculation ends much earlier. I didn't try to understand everything you did, but here's how I'd do it.
$$\begin{vmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} =a_1\underbrace{\begin{vmatrix}b_2 & b_3\\ c_2 & c_3\end{vmatrix}}_{\displaystyle=(\vec B\times\vec C)_1} -a_2\underbrace{\begin{vmatrix}b_1 & b_3\\ c_1 & c_3\end{vmatrix}}_{\displaystyle=-(\vec B\times\vec C)_2} +a_3\underbrace{\begin{vmatrix}b_1 & b_2\\ c_1 & c_2\end{vmatrix}}_{\displaystyle=(\vec B\times\vec C)_3} =\vec A\cdot(\vec B\times\vec C)$$
This is the link to the FAQ post about how to make the math look pretty:

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

If you hit the quote button next to my post, you can see how I did what I did. Same thing with lurflurf's post.

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