So, matrices have always kind of puzzled me, but what puzzles me more has to do with the determinant of a matrix of vectors.

If you have a matrix composed of n-1 n-dimensional vectors and a row of the form [i,j,k,...] (i.e. the unit vectors along each of the axes), the determinant is a vector perpendicular to the ones you started with. (So, if you have the vector <1,2> and put it in a matrix with [i,j] as the other row, you'll get the vector 2i-j or <2,-1>, which is perpendicular to <1,2>. You could also get j-2i or <-2,1> which is also perpendicular.)

You've probably seen this in calculating the cross product. What I want to know is why this works. (It seems to work for n at least up to 6 -- I couldn't be bothered to take the determinant of a 7x7 matrix, but my guess is it will work)

Do any of you have a way to explain why this works?

Thanks!

EDIT: The position of the row [i,j,k,...] might matter -- I'm not sure -- but I'm not about to calculate the determinant of a 3x3 matrix without scratch paper... (It appears as if it works if this row is on top or on the bottom -- and 3x3 is the minimum to test somewhere that isn't the top or bottom...)

## Vectors and Matrix Determinants

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: Vectors and Matrix Determinants

Let's try to find an n×1 vector v

Now we can just worry about finding a matrix inverse, since we would have that [X v

Therefore a vector constructed by your method must be orthogonal to all the other vectors, since if I give you some random vector "w" that is linearly independent from {v

_{n}so that v_{1}^{T}v_{n}= ... = v_{n-1}^{T}v_{n}= 0. I define V = [ v_{1}v_{2}... v_{n-1}] so that we have V^{T}v_{n}= [0 ... 0]^{T}. Now let's make up a vector "w" and matrix "X" so that [V w] and [X v_{n}] are square, and [V w]^{T}[X v_{n}] = I_{n}, where I_{n}is the n×n identity matrix. Expanding this out gives the system of equations V^{T}X = I_{n-1}, V^{T}v_{n}= 0, w^{T}X = 0, w^{T}v_{n}= 1. So far so good.Now we can just worry about finding a matrix inverse, since we would have that [X v

_{n}] = [V w]^{-T}. We can get a v_{n}that works by looking at the last column of [V w]^{-T}. It turns out there is a nifty way to calculate the inverse of a matrix using cofactors. What happens is that the i^{th}element of v_{n}is some constant times (-1)^{i}times the determinant of (V with the i^{th}row deleted). This is the same as the determinant of the matrix you described, times a constant!Therefore a vector constructed by your method must be orthogonal to all the other vectors, since if I give you some random vector "w" that is linearly independent from {v

_{1}, ... , v_{n-1}}, you can just look at the last column of the matrix [V w]^{-T}to get an orthogonal vector that happens to be a constant times the vector you can construct by your method.### Re: Vectors and Matrix Determinants

In addition to cyanoshi's post, I think these 2 videos might help you, at least for the case of 3d vectors:

https://youtu.be/eu6i7WJeinw

https://youtu.be/BaM7OCEm3G0

These 2 videos are part of a series about linear algebra.

https://youtu.be/eu6i7WJeinw

https://youtu.be/BaM7OCEm3G0

These 2 videos are part of a series about linear algebra.

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