Answer :
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.
