# Linear Algebra

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LINALG-MXK4VD

Given the vectors:

\begin{align*} \vec{x}_1&=\begin{pmatrix} 1 \\\ 1\\\ 1 \end{pmatrix} \\\ \vec{x}_2&=\begin{pmatrix} 1 \\\ 0\\\ \!\!-\!1 \end{pmatrix} \\\ \vec{x}_3&=\begin{pmatrix} -1\\\ 0\\\ 1 \end{pmatrix} \\\ \end{align*}

...then $\mathcal{B}=\{\vec{x}_1,\vec{x}_2,\vec{x}_3\}$ is a basis for $\mathbb{R}^3$. If we apply the Gram-Schmidt process to $\mathcal{B}$ to get the basis $\mathcal{B}'=\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$ where we start by letting $\vec{v}_1=\vec{x}_1$ then how many vectors do $\mathcal{B}$ and $\mathcal{B}'$ have in common?

A

0

B

1

C

2

D

3