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ankoles [38]
3 years ago
12

The following statement is either true​ (in all​ cases) or false​ (for at least one​ example). If​ false, construct a specific e

xample to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is​ true, give a justification. If v1 and v2 are in set of real numbers R Superscript 4 and v2 is not a scalar multiple of v1​, then ​{v1​,v2​} is linearly independent. Choose the correct answer below. A. The statement is false. The vector v1 could be a scalar multiple of vector v2. B. The statement is false. The vector v1 could be the zero vector. C. The statement is true. A set of vectors is linearly independent if and only if none of the vectors are a scalar multiple of another vector. D. The statement is false. The vector v1 could be equal to the vector v2.
Physics
1 answer:
Jlenok [28]3 years ago
3 0

Answer:

  • B. The statement is false. The vector v1 could be the zero vector.

Explanation:

Two vectors \vec{v}_1 and \vec{v}_2 are linearly independent if the equation

\alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 = \vec{0}

where \alpha_1 and \alpha_2 are scalars, has only one solution:

\alpha_1 = \alpha_2 = 0.

If there is more than one solution, we say that the vector are linearly dependent.

<h3>Why the answer is B. :</h3>

if \vec{v}_1 = 0 then, \alpha_1 could have any value of the scalar group, as for any scalar \alpha

\alpha * \vec{0} = \vec{0}

So, we get that there is more than one solution.

So, a particular counterexample is:

\vec{v}_1  = (0,0,0,0)

\vec{v}_2 = (1,0,0,0)

as \vec{v}_2 is not an scalar multiple of \vec{v}_2, and the equation

\alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 = \vec{0}

has as solution

\alpha_1 = 2

\alpha_2 = 0

as we can see

2 (0,0,0,0) + 0 (1,0,0,0) = \vec{0}

(2* 0,2 *0,2*0,2*0) +  (0*1,0*0,0*0,0*0) = \vec{0}

(0,0,0,0) +  (0,0,0,0) = \vec{0}

(0+0,0+0,0+0,0+0) = \vec{0}

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