Explain why S 2 1 3 2 1 3 8 4 12 is not a basis for R3 Dete

Explain why S = {(2, 1, -3), (-2, -1, 3), (8, 4, -12)} is not a basis for R^3. Determine whether S = {(8, 7, 6), (0, 7, 6), (0, 0, 6)} is a basis for R^3. If it is, write u = (24, 14, 36) as a linear combination of the vectors in S. Ans: S is a basis for R^3 and u = 3(8, 7, 6) - (0, 7, 6) + 4(0, 0, 6) = (24, 14, 36). Determine whether S = {(0, 0, 0), (1, 3, 6). (8, 1, -3)} is a basis for R^3. If it is, write u = (0, 30, 60) as a linear combination of the vectors in S. Ans: S is not a basis for R^3.

Solution

11.   To ascertain whether the vectors in S form a basis for R³, we need to check whether these vectors are linearly independent and span R³ . Let A be the matrix with these vectors as columns. We will reduce A to its RREF as under:

Multiply the 1st row by ½

Add -1 times the 1st row to the 2nd row

Add 3 times the 1st row to the 3rd row

Then the RREF of A is

1

-1

4

0

0

0

0

0

0

This shows that (-2,-1,3) = -(2,1,-3) and (8,4,-12)= 4(2,1,-3). ( it is apparent even otherwise).

Hence the vectors in S are neither linearly independent and nor do they span R³. Hence S is not a basis for R³.

12. To ascertain whether the vectors in S form a basis for R³, we need to check whether these vectors are linearly independent and span R³ . Let A be the matrix with these vectors, and u as columns. We will reduce A to its RREF as under:

Multiply the 1st row by 1/8; Add -7 times the 1st row to the 2nd row

Add -6 times the 1st row to the 3rd row; Multiply the 2nd row by 1/7

Add -6 times the 2nd row to the 3rd row; Multiply the 3rd row by 1/6

Then the RREF of A is

1

0

0

3

0

1

0

-1

0

0

1

4

This shows that the vectors in S are linearly independent and span R³. Therefore, S is a basis for R³.Further, u = 3(8,7,6)-1(0,7,6)+4(0,0,6).

13. To ascertain whether the vectors in S form a basis for R³, we need to check whether these vectors are linearly independent and span R³ . Let A be the matrix with these vectors and u as columns. . We will reduce A to its RREF as under:

Add -3 times the 1st row to the 2nd row

Add -6 times the 1st row to the 3rd row

Multiply the 2nd row by -1/23

Add 51 times the 2nd row to the 3rd row

Multiply the 3rd row by -23/150

Add 30/23 times the 3rd row to the 2nd row

Add -8 times the 2nd row to the 1st row   

Then the RREF of A is

0

1

0

0

0

0

1

0

0

0

0

1

This shows that the vectors in S are not linearly independent and do not span R³. Therefore, S is not a basis for R³.Further, u cannot be expressed as a linear combination of the vectors in S.

1	-1	4
0	0	0
0	0	0