Assume u4 is not a linear combination of {u1,u2,u3}

select the best statement.

a. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.
b. {u1,u2,u3,u4}could be a linearly independent or linearly dependent set of vectors depending on the vector space chosen.
c. {u1,u2,u3,u4}is always a linearly dependent set of vectors.
d. {u1,u2,u3,u4} is never a linearly dependent set of vectors.
e. {u1,u2,u3,u4}is a linearly dependent set precisely when {u1,u2,u3} is a linearly dependent set.
f. none of the above

A. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.

Step-by-step explanation:

Given that u4 is not a linear combination of {u1,u2,u3}

This means there is no possibility to write u4 = au1+bu2+cu3 for three scalars a,b,and c.

This gives that

This implies that these four vectors are not linearly dependent but linearly independent.

Hence option a is right.

A. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

80

step-by-step explanation: