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klimenkov

A tutorial for @RolyPoly. Linear independence of the vectors. Check if the vectors \(\vec{v_1}=(1,0,1), \vec{v_2}=(2,1,3), \vec{v_3}=(1,1,2)\) are linearly independent?

  • one year ago
  • one year ago

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  1. RolyPoly
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    \[\left[\begin{matrix}1 & 2 & | & 1 \\ 0 & 1 &|& 1 \\ 1&3&|&2\end{matrix}\right]\]\[->\left[\begin{matrix}1 & 2 & | & 1 \\ 0 & 1 &|& 1\\ 0&1&|&1\end{matrix}\right]\]\[->\left[\begin{matrix}1 & 2 & | & 1 \\ 0 & 1 &|& 1\\ 0&0&|&0\end{matrix}\right]\] => Linearly dependent :|

    • one year ago
  2. klimenkov
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    We will use a definition of the linear dependence. Start from making the linear combination of \(\vec{v_1},\vec{v_2},\vec{v_3}\): \(c_1\cdot\vec{v_1}+c_2\cdot\vec{v_2}+c_3\cdot\vec{v_3}\) And we will try to check if there are such \(c_1,c_2,c_3\) to make this combination equal to zero. \(c_1\cdot\left(\begin{matrix}1\\0\\1\end{matrix}\right)+c_2\cdot\left(\begin{matrix}2\\1\\3\end{matrix}\right)+c_3\cdot\left(\begin{matrix}1\\1\\2\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\end{matrix}\right)\). That means that we have a system of the linear equation with the right part equal to zero: \(A\vec{x}=0\) Where \(A=\left(\begin{matrix} 1&2&1\\ 0&1&1\\ 1&3&2 \end{matrix}\right), \vec{x}=\left(\begin{matrix}c_1\\c_2\\c_3\end{matrix}\right)\). Using Cramer's rule we compute the determinant: \(\left|\begin{matrix} 1&2&1\\ 0&1&1\\ 1&3&2 \end{matrix}\right|=0\) We have that the right side of the matrix equation \(A\vec{x}=0\) is equal to zero and the determinant of the matrix \(A\) is equal to zero. That means that there is non-trivial solution for \(\vec{x}\). We found that there are so \(c_1,c_2,c_3\) not all are equal to zero. According to the definition of the linear independence of the vector, we can say that \(\vec{v_1},\vec{v_2},\vec{v_3}\) are linearly dependent.

    • one year ago
  3. RolyPoly
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    We have that the right side of the matrix equation \(A\vec{x} =0\) is equal to zero and the determinant of the matrix A is equal to zero. ^ Okay. That means that there is non-trivial solution for \(\vec{x}\) ^ Not okay.

    • one year ago
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