Word Count : 674

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To show that the vectors \[ \vec{u}_1 = \begin{bmatrix} 2 \\ 0 \\ -3 \end{bmatrix}, \quad \vec{u}_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \vec{u}_3 = \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} \] are linearly independent, we need to check whether the only solution to the equation \[ c_1 \vec{u}_1 + c_2 \vec{u}_2 + c_3 \vec{u}_3 = \vec{0} \] is the trivial solution \( c_1 = c_2 = c_3 = 0 \). ### Step 1: Set Up the Homogeneous System \[ c_1 \begin{bmatrix} 2 \\ 0 \\ -3 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] This gives the system of equations: \[ 2c_1 + c_2 + c_3 = 0 \] \[ 0c_1 + 1c_2 + 7c_3 = 0 \] \[ -3c_1 + 1c_2 + 2c_3 = 0 \] ### Step 2: Convert to Augmented Matrix \[ \begin{bmatrix} 2 & 1 & 1 & | ________ __________ _____ _______ _______ _____.
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