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Dalam ruang vektor ๐‘‰, untuk setiap vektor ๐‘ขโƒ—, harus ada suatu vektor โˆ’๐‘ขโƒ—dalam ๐‘‰ sehingga....In the vector space ๐‘‰, for every vector ๐‘ขโƒ—, there must exist a vector โˆ’๐‘ขโƒ— in ๐‘‰ such that....A๐‘ขโƒ—โˆ’๐‘ขโƒ—=0โƒ—B๐‘ขโƒ—+(โˆ’๐‘ขโƒ—)=0โƒ—C๐‘ขโƒ—โ‹…(โˆ’๐‘ขโƒ—)=โˆ’1D๐‘ขโƒ—ร—(โˆ’๐‘ขโƒ—)=0โƒ—

Question

Dalam ruang vektor ๐‘‰, untuk setiap vektor ๐‘ขโƒ—, harus ada suatu vektor โˆ’๐‘ขโƒ—dalam ๐‘‰ sehingga....In the vector space ๐‘‰, for every vector ๐‘ขโƒ—, there must exist a vector โˆ’๐‘ขโƒ— in ๐‘‰ such that....A๐‘ขโƒ—โˆ’๐‘ขโƒ—=0โƒ—B๐‘ขโƒ—+(โˆ’๐‘ขโƒ—)=0โƒ—C๐‘ขโƒ—โ‹…(โˆ’๐‘ขโƒ—)=โˆ’1D๐‘ขโƒ—ร—(โˆ’๐‘ขโƒ—)=0โƒ—

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Solution

Sure, let's analyze the given options step by step to determine which one correctly completes the statement about vector spaces.

In a vector space V V , for every vector uโƒ— \vec{u} , there must exist a vector โˆ’uโƒ— -\vec{u} in V V such that:

A. uโƒ—โˆ’uโƒ—=0โƒ— \vec{u} - \vec{u} = \vec{0}

This statement is true because subtracting a vector from itself results in the zero vector, 0โƒ— \vec{0} . However, this is not the standard definition of the additive inverse in vector spaces.

B. uโƒ—+(โˆ’uโƒ—)=0โƒ— \vec{u} + (-\vec{u}) = \vec{0}

This statement is the correct definition of the additive inverse in vector spaces. For every vector uโƒ— \vec{u} , there exists a vector โˆ’uโƒ— -\vec{u} such that their sum is the zero vector, 0โƒ— \vec{0} .

C. uโƒ—โ‹…(โˆ’uโƒ—)=โˆ’1 \vec{u} \cdot (-\vec{u}) = -1

This statement is not generally true. The dot product of a vector and its negative is not necessarily โˆ’1-1; it depends on the magnitude of the vector. In fact, uโƒ—โ‹…(โˆ’uโƒ—)=โˆ’โˆฅuโƒ—โˆฅ2 \vec{u} \cdot (-\vec{u}) = -\|\vec{u}\|^2 , which is not always โˆ’1-1.

D. uโƒ—ร—(โˆ’uโƒ—)=0โƒ— \vec{u} \times (-\vec{u}) = \vec{0}

This statement is true in three-dimensional space because the cross product of any vector with itself (or its negative) is the zero vector. However, this is not the standard definition of the additive inverse in vector spaces.

Therefore, the correct answer is:

B. uโƒ—+(โˆ’uโƒ—)=0โƒ— \vec{u} + (-\vec{u}) = \vec{0}

This option correctly completes the statement about the existence of an additive inverse in a vector space.

This problem has been solved

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