Get personalized and accurate responses to your questions with IDNLearn.com. Ask your questions and receive comprehensive, trustworthy responses from our dedicated team of experts.
Sagot :
The set of all 3x1 matrices of the form \(\begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix}\), denoted as \(V\), does not form a vector space because it fails to satisfy one of the vector space axioms, specifically closure under scalar multiplication.
To demonstrate:
1. **Closure under Scalar Multiplication**: For \(V\) to be a vector space, multiplying any matrix in \(V\) by a scalar should still result in a matrix that belongs to \(V\).
Consider a matrix \( \mathbf{v} = \begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix} \) in \(V\). Let's perform scalar multiplication:
Let \( \alpha \) be a scalar. Then \( \alpha \mathbf{v} = \alpha \begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix} = \begin{bmatrix} \alpha x \\ \alpha(2z - y) \\ 3\alpha \end{bmatrix} \).
Now, check if \( \alpha \mathbf{v} \) still fits the form \( \begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix} \):
- The first entry of \( \alpha \mathbf{v} \) is \( \alpha x \), which is of the form \( x \). This satisfies the first entry condition.
- The second entry of \( \alpha \mathbf{v} \) is \( \alpha(2z - y) \), which is \( 2\alpha z - \alpha y \). This must match \( 2z - y \) for all \( \alpha \), not just a specific value.
To demonstrate:
1. **Closure under Scalar Multiplication**: For \(V\) to be a vector space, multiplying any matrix in \(V\) by a scalar should still result in a matrix that belongs to \(V\).
Consider a matrix \( \mathbf{v} = \begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix} \) in \(V\). Let's perform scalar multiplication:
Let \( \alpha \) be a scalar. Then \( \alpha \mathbf{v} = \alpha \begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix} = \begin{bmatrix} \alpha x \\ \alpha(2z - y) \\ 3\alpha \end{bmatrix} \).
Now, check if \( \alpha \mathbf{v} \) still fits the form \( \begin{bmatrix} x \\ 2z - y \\ 3 \end{bmatrix} \):
- The first entry of \( \alpha \mathbf{v} \) is \( \alpha x \), which is of the form \( x \). This satisfies the first entry condition.
- The second entry of \( \alpha \mathbf{v} \) is \( \alpha(2z - y) \), which is \( 2\alpha z - \alpha y \). This must match \( 2z - y \) for all \( \alpha \), not just a specific value.
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.