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For which values of $k$ is the set of vectors $V=\{v_1,v_2,v_3\}$, where $$v_1=(2,1,-3),\quad v_2=(k-8,-3,2k+5),\quad v_3=(5,2,k-7)$$ linearly independent?

A set is L.I. if we have $$\lambda_1v_1+\lambda_2v_2+\lambda_3v_3= 0$$ by Gaussian elimination I have reduced the matrix of coefficients for each $\lambda_i$ by Gaussian elimination to $$\begin{pmatrix}1&1-2k&0\\0&k-2&1\\0&0&k-3\end{pmatrix}$$ note I have switched row 1 and row 2. Now can I just deduce that this set is L.I. If $k\not=3$, $k\not=2$ and $k\not=1/2$.

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  • $\begingroup$ They are L.I. if the determinant of that matrix is $\neq 0$. It's an upper triangular matrix, the determinant is just the product of the $a_{i,i}$'s. Therefore $k\neq 2$ and $k\neq 3$. I'll suggest to check the Gaussian elimination, just to be sure. $\endgroup$ – Diego Robayo Feb 19 '15 at 22:37
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Yes this is correct, though $k=1/2$ does not matter. Only the diagonal in your elimination controls the rank.

You could also note that $$ \det \begin{bmatrix} 2 & 1 & -3\\ k-8 & -3 & 2\,k + 5 \\ 5&2& k-7 \end{bmatrix} = -(k-3)(k-2) $$ Hence the vectors are linearly independent if and only if $k\neq 3$ and $k\neq 2$.

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If your Gaussian elimination calculation is correct, then the only way the original matrix can be non-invertible (hence the original vectors are not linearly independent) is if one of the diagonal entries in the Gaussian elimination matrix is 0. So that would mean $k \neq 2$ and $k \neq 3$, but $k = 1/2$ would be ok.

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