Use cross product formula in R^4 to obtain orthogonal vector

In summary, the matrix A = \begin{bmatrix} 1 & 4 & -1 & 2 \\ 0 & 1 & \phantom{-}0 & -1 \\ 2 & 9 & -2 & 2\end{bmatrix} has an inconsistent matrix multiplication that yields x1+ 4x2 - 3x3 + 2x4 =0 and x2 + x4 = 0. This corresponds to a matrix A=\begin{bmatrix} 1 & 4 & -3 & 2 \\ 0 & 1 & \phantom{-}0 & 1 \\ 2 & 9 & -2 & 2\end{bmatrix} which is not the matrix as shown in the OP.
  • #1
DryRun
Gold Member
838
4
Use cross product formula in R^4 to obtain a vector that is orthogonal to rows of A

Please help with first part and check if i answered the questions correctly.

The matrix A =
1 4 -1 2
0 1 0 -1
2 9 -2 2


1. Use cross product formula in R^4 to obtain a vector that is orthogonal to the rows of A.

I multiply the matrix A by a 4x1 matrix X and equate to 0.

Matrix X =
x1
x2
x3
x4

Then i get:
x1+ 4x2 - 3x3 + 2x4 =0
x2 + x4 = 0
2x1 + 9x2 - 2x3 + 2x4 =0


x1 = -4x2 + 3x3 - 2x4
x2 = - x4
x3 = x3
2x4 = -2x1- 9x2 + 2x3

And then... I'm stuck.

2. How is this vector related to the null space of A?
My answer: That vector is perpendicular to the null space of A.
Is this correct and is there another way to put it?
 
Last edited:
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  • #2
sharks said:
The matrix A =
1 4 -1 2
0 1 0 -1
2 9 -2 2

...

x1+ 4x2 - 3x3 + 2x4 =0
x2 + x4 = 0
2x1 + 9x2 - 2x3 + 2x4 =0
Your first two equations are inconsistent with the matrix you posted.

The title of the thread doesn't make much sense. There is no cross product formula in R^4. Do you mean the formula for the product of a matrix and a vector? (Which you did use, if erroneously.)

You will end up with three equations in four unknowns. The second row of A let's you quickly eliminate one of x1, x2, x3, and x4. That will leave you with two equations in three unknowns. See if you can get to, and possibly beyond, this point.
 
  • #3
OK, i will get to it for the first part. But is the second part correct?
 
  • #4
What second part?
 
  • #5
sharks said:
2. How is this vector related to the null space of A?
My answer: That vector is perpendicular to the null space of A.
Is this correct and is there another way to put it?

This part.
 
  • #6
The vector is in the null space of A, not perpendicular to it.
 
  • #7
I'm sorry but i haven't understood anything from your replies. It's obvious that you haven't even taken the time to read my question properly. You missed my second question despite using bold for the questions, and you haven't been helpful either with the first part of the question. The question is as it is. The title is based on part of the first question. The entire question itself (as well as my attempts) is in the first post.
Not being ungrateful here, but your replies have made this whole troubleshooting process even more confusing, if that's even possible.

I have fixed the title to make it match the first question.
 
  • #8
What is unclear about say that you did the matrix multiplication wrong? I gave you a hint for how to go further after you get that multiplication correct. I'll help you more if you get stuck.

What is unclear about saying that you were wrong in saying that "That vector is perpendicular to the null space of A"? For crying out loud, I out and out gave you the answer to that part of the question.
 
  • #9
For part 1, I think you're actually being asked to calculate
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} & \hat{w} \\
1 &4 &-1 &2 \\
0 &1 &0 &-1 \\
2 &9 &-2 & 2
\end{vmatrix}
 
  • #10
I took the first part as meaning finding a 4-vector such that it is orthogonal to each of the three rows of the given 3x4 matrix. I take the title of the thread as meaning that any 4-vector that satisfies these conditions will suffice; all that is needed is one (as opposed to a a general expression that describes all such 4-vectors).

Since the OP hasn't come back in a while, the problem with the first part is that the matrix multiplication does not jibe with the matrix. Given

[tex]A=\begin{bmatrix}
1 & 4 & -1 & \phantom{-}2 \\
0 & 1 & \phantom{-}0 & -1 \\
2 & 9 & -2 & \phantom{-}2\end{bmatrix}[/tex]

and [itex]x=[x_1\,\,x_2\,\,x_3\,\,x_4]^T[/itex], the matrix multiplication [itex]Ax[/itex] yields

[tex]
\begin{matrix}
x_1 &+& 4x_2 &-& x_3 &+& 2x_4 &=& 0 \\
&&\phantom{1}x_2 && &-& \phantom{1}x_4 &=& 0 \\
2x_1 &+& 9x_2 &-& 2x_3 &+& 2x_4 &=& 0
\end{matrix}[/tex]

The OP came up with
x1+ 4x2 - 3x3 + 2x4 =0
x2 + x4 = 0
2x1 + 9x2 - 2x3 + 2x4 =0

This corresponds to a matrix

[tex]A=\begin{bmatrix}
1 & 4 & -3 & 2 \\
0 & 1 & \phantom{-}0 & 1 \\
2 & 9 & -2 & 2\end{bmatrix}[/tex]

So which is correct, the matrix as shown in the OP, or the matrix as inferred from the OP's matrix multiplication?
 
  • #11
I don't know either. The OP needs to clarify which matrix is correct and exactly what the problem statement was. I just inferred from the mention of the "cross product" that the OP was meant to use the generalization to R4 involving the determinant.
 
  • #12
The matrix multiplication that i did is wrong. By the way, how do i type the matrices in here as neatly as you did?
 

Related to Use cross product formula in R^4 to obtain orthogonal vector

1. What is the cross product formula in R^4?

The cross product formula in R^4 is a mathematical operation used to find a vector that is perpendicular to two given vectors in a four-dimensional space. It is also known as the vector product or outer product.

2. How do you use the cross product formula in R^4?

To use the cross product formula in R^4, you need to have two given vectors in the four-dimensional space. Then, you can apply the formula, which involves multiplying the components of the vectors and applying some algebraic operations to obtain the resulting orthogonal vector.

3. Why is the cross product formula useful in R^4?

The cross product formula in R^4 is useful because it allows us to find a vector that is perpendicular to two given vectors, which is a fundamental concept in vector algebra. This is important in many applications, such as finding the normal vector to a plane or calculating torque in physics.

4. Can the cross product formula be used in other dimensions besides R^4?

Yes, the cross product formula can be used in any odd-dimensional space. In fact, it is most commonly used in three-dimensional space (R^3), where it is frequently used to find the area of a parallelogram formed by two given vectors.

5. Are there any limitations to using the cross product formula in R^4?

One limitation of the cross product formula in R^4 is that it can only be used to find a perpendicular vector in that specific four-dimensional space. It cannot be applied to vectors in a different number of dimensions. Additionally, the result of the cross product may not always be unique, as there may be multiple orthogonal vectors that satisfy the given conditions.

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