$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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proof
$\text{det}(v_1, \dots, v_n)=0$ if and only if $\{v_1, \dots, v_n\}$ is linearly dependent, and a matrix $A$ with columns $v_1, \dots, v_n$ is not invertible.

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$\text{det}(v_1, \dots, v_n)=0$ if and only if $\{v_1, \dots, v_n\}$ is linearly dependent, and a matrix $A$ with columns $v_1, \dots, v_n$ is not invertible.

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concepts
If
Only If
proof
FullPage
result
Concepts
If
Only If
proof
$\text{det}(v_1, \dots, v_n)=0$ if and only if $\{v_1, \dots, v_n\}$ is linearly dependent, and a matrix $A$ with columns $v_1, \dots, v_n$ is not invertible.

Concepts

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If

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Only If

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Proof

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$\text{det}(v_1, \dots, v_n)=0$ if and only if $\{v_1, \dots, v_n\}$ is linearly dependent, and a matrix $A$ with columns $v_1, \dots, v_n$ is not invertible.

Concepts

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If

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Only If

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Proof

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concepts
If
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proof