[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

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)