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[SOLVED] 14.1 find a vector v that will satisfy the system


Well-known member
Jan 31, 2012
View attachment 9050
ok I think I got (a) and (b) on just observation

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


Well-known member
MHB Math Helper
Jan 29, 2012
A basis for a vector space of dimension n has three properties:
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, two vectors are dependent if and only if one is a multiple of the other. Here, it is sufficient to observe that [tex]\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}[/tex] is not a multiple of [tex]\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}[/tex].)

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. IF the three vectors were dependent then v would be a linear combination of the other two. In particular, taking the the coefficients to be 1, the sum of the two given vectors is [tex]\{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}+
\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.)


Well-known member
Jan 31, 2012
ok I can see that the multiplicity of the matrix's would be an easier way to check for linearity
but they seem to use the augmented matrix in the examples

thank you for the expanded explanation that was a great help