[10 points] Let v1 = (−2, −7, 2), v2 = (5, 1, −5), v3 = (−4, −9, 4), v4 = (−3, −6, 3), v5 = (1, 2, −1). Find a subset of {v1, v2, v3, v4, v5} that forms a basis for the subspace of R 3 spanned by those five vectors. Express each non-basis vector as a linear combination of basis vectors

A basis is {v1,v4}

v2 = 3 v1 -11/3 v4

v3 = 1/3 v1 + 10/9 v4

v5 = -1/3 v4

Step-by-step explanation:

Lets triangulate the matrix that has the elements v1, v2, v3, v4 and v5 as their rows

Lets divide each row by its first element, which is possible because it is different than 0

Thus v5 = v4 / -3 = -1/3 * v4. We can take it out. Also, we will take each row except the first one and substract the value of the first row from them, obtaining

Note that now, from the second row onwards, each row is a multiple of the other. So we can remove every row except 2 of them, for example the first one and the fourth one

We can form a base of the subspace of R³ spanned by v1,v2,v3,v4,v5 with {v1,v4}. Also, we have

v5 = -1/3*v4

(v1/-2-v3/-4)*1.5 = (v1/-2-v4/-3)*1.25, hence

0.25v1/-2 + 1.25 v4/-3 = 1.5 v3/-4

Thus

v3 = -8/3 *0.25 v1/-2 - 8/3*1.25 v4/-3 = 1/3 v1 + 10/9 v4

In a similar way,

(v1/-2-v2/5)*1.5 = (v1/-2-v4/-3)*3.3

Thus

v2 = 10/3 * (-1.8v1/-2 + 3.3v4/-3) = 3 v1 -11/3 v4

b

2

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b

b

3

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7

b

2

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14

b

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6

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−

3

b

3

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7

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14

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6

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step-by-step explanation:

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.