- #1

karush

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ok I think I got (a) and (b) on just observation

but (c) doesn't look like x,y,z will be intergers so ?

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- #1

karush

Gold Member

MHB

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ok I think I got (a) and (b) on just observation

but (c) doesn't look like x,y,z will be intergers so ?

- #2

HallsofIvy

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1) The vectors are independent

2) The vectors span the space

3) There are n vectors in the space.

Further, any two of those imply the third.

The simplest way to answer the first question is that a basis for a three dimensional vector space, such as [tex]R^3[/tex], must contain three vectors, not two.

For the second, rather than using matrices, I would write

[tex]a\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}+ b\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}= \begin{bmatrix}a+ b \\ a+ 2b \\ a+ 3b\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}[/tex]

so we must have a+ b= 0, a+ 2b= 0, and a+ 3b= 0. Subtracting the first equation from the second we have b= 0 and then a+ b= a+ 0= 0 so a= 0. The coefficients are a= b= 0 so the vectors are independent.

(Actually,

There are, of course, infinitely many correct answers to

(c) Find a vector, v, such that [tex]\{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}, v\}[/tex] is a basis for [tex]R^3[/tex]. Since those are three vectors it sufficient to find v such that the three vectors are independent or such that they span [tex]R^3[/tex]. You appear to have chosen to find v such that the three vectors are independent but you haven't finished the problem.

\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}= \begin{bmatrix}2 \\ 3 \\ 4\end{bmatrix}

. To find a vector, v, that is NOT a linear combination, just change one of those components. Change, say, the "4" to "5": [tex]v= \begin{bmatrix}2 \\ 3 \\ 5 \end{bmatrix}[/tex]. The first two components are from "u+ v" while the third is not so v is not any linear combination.

(Since these are 3-vectors, you could also take v to be the "cross product" of the two given vectors. The cross product of two vectors is perpendicular to both so independent of them.)

(Since these are 3-vectors, you could also take v to be the "cross product" of the two given vectors. The cross product of two vectors is perpendicular to both so independent of them.)

- #3

karush

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but they seem to use the augmented matrix in the examples

thank you for the expanded explanation that was a great help

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