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$\vec{a}=\hat{i}-\hat{k}$ ,

$\vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and

$\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$ , then

$\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{bmatrix}$ depends on

1
2
3
4

Given:

$\vec{a} = \hat{i} - \hat{k}, \quad \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$

Form the matrix:

$M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix}$

Find the determinant:

$\det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1$

Since the determinant is constant and non-zero, the vectors are linearly independent.

$\boxed{\text{The matrix does not depend on } x \text{ or } y}$

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