# Linear Combination in Vector

**Definition.**

A vector w is called a linear combination and vectors v

_{1}, v_{2}, ..., v_{r}if the vector can be expressed in form
w = k

_{1}v_{1}+ k_{2}v_{2}+ ... + k_{r}v_{r}
Where k

_{1}, k_{2}, ..., k_{r}is scalar.**Example**

**1**

**.**

Review the vectors

**u**= (1, 2, -1) and**v**= (6, 4, 2) on R^{3}. Show that**w**= (9, 2, 7) is a linear combination**u**and**v**and**w’**= (4, -1, 8) is not a linear combination**u**and**v**.**Solution:**

So that w is a linear combination of

**u**and**v**, there must be a scalar of k_{1}and k_{2}so that
w = k

_{1}**u**+ k_{2}**v**
that is

(9, 2, 7) = k

_{1}(1, 2, -1) + k_{2}(6, 4, 2)
or

(9, 2, 7) = (k

_{1}+ 6k_{2}, 2k_{1}+ 4k_{2}, -k_{1}+ 2k_{2})
Equation of the
appropriate components gives

k

_{1}+ 6k_{2}= 9
2k

_{1}+ 4k_{2}= 2
- k

_{1}+ 2k_{2}= 7
By solving this system it will produce k

_{1}= -3, k_{2}= 2 so that
w = -3

**u**+ 2**v**

likewise, so that

**w’**is a linear combination of**u**and**v**, there must be a scalar of k_{1}and k_{2}so that**w’**= k_{1}**u**+ k_{2}**v**, i.e.
(4, -1, 8) = k

_{1}(1, 2, -1) + k_{2}( 6, 4, 2)
or

(4, -1, 8) = (k

_{1}+ 6k_{2}, 2k_{1}+ 4k_{2}, -k_{1}+ 2k_{2})
Equation of the
appropriate components gives

k

_{1}+ 6k_{2}= 4
2k

_{1}+ 4k_{2}= -1
- k

_{1}+ 2k_{2}= 8
This system of equations
is inconsistent, so there are no such scalars. As a consequence,

**w'**is not a linear combination of**u**and**v**.**Definition.**

If v

_{1}, v_{2}, ..., v_{r}are vectors in vector space**V**and if each vector in**V**can be expressed as a linear combination v_{1}, v_{2}, ..., v_{r}we say that these vectors span**V**.**Example 2**.

The vectors

**i**= (1, 0, 0),**j**= (0, 1, 0) and**k**= (0, 0, 1) span R^{3}because each vector (a, b, c) on R^{3}can be written as
(a, b, c) = a

**i**+ b**j**+ c**k**
Which is a linear
combination

**i, j, k**.**Example 3.**

Determine whether

**u**= (1, 1, 2),**v**= (1, 0, 1) and**w**= (2, 1, 3) span R3?
Solution:

We must determine whether
any vector

**b**= (b_{1}, b_{2}, b_{3}) in R^{3}can be expressed as a linear combination.**b**= k

_{1}

**u**+ k

_{2}

**v**+ k

_{3}

**w**

From vectors

**u**,**v**, and**w**. By expressing this equation in the components it will give
(b

_{1}, b_{2}, b_{3}) = k_{1}(1, 1, 2) + k_{2}(1, 0, 1) + k_{3}(2, 1, 3)
or

(b

_{1}, b_{2}, b_{3}) = k_{1}+ k_{2}+ 2k_{3}, k_{1}+ k_{3}, 2k_{1}+ k_{2}+ 3k_{3}_{}

By equating the
corresponding components giving

k

_{1}+ k_{2}+ 2k_{3 }= b_{1}
k

_{1}+ k_{3 }= b_{2}
2k

_{1}+ k_{2}+ 3k_{3 }= b_{3}_{}

This system will be
consistent for all values b

_{1}, b_{2}, and b_{3}if and only if the matrix coefficients,**u**,

**v**and

**w**do not stretch R

^{3}.

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